07-20-12

07-20-12

Exercise in confidence interval for proportion, variance and standard deviation

1.In a recent survey of 150 households, 54 has central air conditioning.  Find the p -hat and the q-hat, where p-hat is the proportion of households that have centralized air conditioning.

2. A sample of 500 nursing applicants included 60 from men.  Find the 90%confidence interval of the true proportion of men who applied to the nursing program.

3. A survey of 200,000 boat owners found that 12% of the pleasure boats were named serenity.  Find the 95 % confidence interval of the true proportion of boats named serenity.

BY: JOREY MARK MILLAMENA

07-13-12

07-13-12

Exercise in confidence intervals for the mean (standard deviation unknown or sample size less than 30)

1. Find the ta/2 value for 95% confidence interval when the sample size is 22.

2. Ten randomly selected automobiles were stopped and the tread from right of the tire was measured.  The mean was 0.32 inch and the standard deviation was 0.08 inch.  Find the 95%confidence interval of the mean depth.  Assume that the variable is normally distributed.

3.  A recent study of 28 city residents showed that the mean of the time they had lived at their present address was 9.3 years.  The standard deviation of the sample was 2 years.  Find the 90% confidence interval of the true mean.

BY:JOREY MARK MILLAMENA

07-06-12

 07-06-12

Exercise in confidence intervals for the mean (standard deviation known or sample size greater than or equal to 30)

1.      Find each.

a.       Za/2 for the 99% confidence interval

b.      Za/2 for the 98% confidence interval

c.       Za/2 for the 95% confidence interval

d.      Za/2 for the 90% confidence interval

e.       Za/2 for the 94% confidence interval

2.       A university dean wishes to estimate the average number of hours hi part-time instructors teach per week.  The standard deviation from a previously study is 2.6 hours.  How large a sample must be selected if he wants to be 99% confident of finding whether the true mean differs from the sample mean buy 1 hour?

BY: JOREY MARK MILLAMENA

07-23-12 ( JULY 23, 2012); Z- test and T- test for the Mean

On July 23 we III- Gold students were taught a new lesson which is the Z-test and T- test for the Mean.
In our lesson:

•       Hypotheses are tested using a statistical test based on this general formula which is the:

Test value= (observed value-expected value)/ standard error

Observed value: the computed sample

Expected value: the parameter expected when the null hypothesis were true

Standard error: the error of the statistics being tested

• Formula for Z-test and T-test

•  Z-test for the mean is used when (n) is greater or equal to 30 or the population mean is normally distributed and the standard deviation is known.
• T-test for the mean is used when (n) is less than 30 and when the standard deviation is not known.
• Four possible outcomes and the summary statement for each situation:

When Claim is the Null Hypothesis

-          Reject The Null Hypothesis; There is enough evidence to reject the claim

-          Do not Reject the Null Hypothesis; There is not enough evidence to reject the claim

When Claim is the Alternative Hypothesis

-          Reject The Null Hypothesis; There is enough evidence to support the claim

-          Do not Reject the Null Hypothesis; There is not enough evidence to support the claim

•   P-Value Method for the Hypothesis Testing – is the actual area under the standard normal distribution curve, representing the probability of a particular sample statistics or a more extreme sample statistic occurring if the null hypothesis is true.
• Steps in Hypothesis testing  problems in P-Value Method

Step1. State the hypothesis and identify the claim

Step2. Compute the test value

Step3. Find the P-Value

Step 4.Make the decision

Step5. Summarize the Results

•   Decision rule when using a P-Value

If P-Value is less than or equal to alpha, reject the null hypothesis.

If P-Value is greater than the alpha, do not reject the null hypothesis.

By: Carl Joel E. Palma III-Gold

July 25,2012-Lesson on z-test for the Mean

I can’t really digest all of our previous lessons because just these weeks, I always have my early morning and late afternoon migraines.. Just like in my advanced Stat., we have our lesson on z-test and t-test for the mean. These lessons need so much attentiveness in order for me to analyzed and have the correct decision in my answer such as rejecting or not rejecting the null hypotheses and on having my conclusion.

I just don’t really mind my migraine because I need to fully understand our lesson.

Here is one of the problems that we had:

A researcher reports that the average salary of assistant professors is more than $42,000. A sample of 30 assistant professors has a mean salary of $43,260. At  α=0.05, test the claim that assistant professors earn more than $42,000 a year. The standard deviation of the population is $5,230.

In order for us to answer problems like this, we should need to follow these 5 steps:

Step 1: State the hypotheses and identify the claim.

Null Hypotheses (H₀):    µ ≤ $42,00     and

Alternative Hypotheses (H₁):      µ > $42,000 (claim)

Step 2: Find the critical value. Since α=0.05 and the test is a right-tailed test, the critical value is z=1.65.

Step 3: Compute the test value.

 Step 4: Make the decision. Since the test value, 1.32 is less than the critical value 1.65, and is not in the critical region, the decision is “Do not reject the null hypothesis”.

Step 5: Summarize the results. There is not enough evidence to support the claim that assistant professors earns more on average than $42,000 a year.

Even though the sample mean, $43,000, is higher than the hypothesized population mean of $42,000, it is not significantly higher. Hence, the difference may due to chance.

It should be noted that when the null hypothesis is not rejected, it cannot be accepted as true. There is merely not enough evidence to say that it is false. This guideline may sound a little confusing, but the situation is analogous to a jury trial. The verdict is either guilty or not guilty and is based on the evidence presented. If a person is judged not guilty, it does not mean that the person is proved innocent; it only means that there is enough evidence to reach the guilty verdict.

By: Kent Spencer M. Mendez

Four Possible Outcomes and the Summary for each Situation-July 25,2012

 

By: Kent Spencer M. Mendez

III-Gold

July 26, 2012- Discussing of finding the test value where n< 30.

    Last Thursday, Ms. Macatigos discuss further about finding the test value where n is less than 30. She said that the formula for the test value where n <30 is just the as the formula where n > 30. She told as that the formula is t = sample mean- hypothesized population mean/ sample deviation / square root of n or t = x̅ - μ /s/n.

   For Example using the P-Value Method:

           

          A physician claims that joggers maximal volume oxygen uptake is greater than the average of all adults. A sample of 15 joggers has a mean of 40.6 milliliters per kilograms and a standard deviation of 6ml/kg. If the average of all adults is 36.7 ml/kg, is there enough evidence to support the physicians claim at a= 0.05?

Solution:

     STEP 1:State the Hypothesis

                      Null hypothesis = m ≤ 36.7 and alternative hypothesis = m >  36.7

    STEP 2: Compute the test value

                     t= 40.6-36.7 / 6 / square root of 15 = 2.527

    STEP 3: Find the P-Value.

                     Looking across the row with d.f = 14 in Table F, 2.527 falls between

                       2.145 and 2.624. Therefore, 0.01 < P-Value < 0.025

    STEP 4: Make the decision:

                      Reject the null hypothesis since the P- Value < 0.05 ( P-Value< a)

    STEP 5: Summarize the results

                      There is enough evidence to support the claim that the joggers maximal volume oxygen uptake is greater than 36.7 ml/kg.

          

               Hope you agree with the solution and also understand the discussion

       After the short discussion, our teacher gave us some sort of test about the topic to see if really we have learned something and if we are really listening.

By: Dustin Joshua Esquia III- Gold

July 25, 2012. T-test and Z-test for the mean, video.

T-test and z-test for the Mean

 

By: Tito Nuevacobita Jr. III-Gold

Z-test and t-test for the Mean; July 25, 2012

                                                        Introduction to the z and t-tests



                  Z-test and t-test are basically the same; they compare between two means to suggest whether both samples come from the same population. There are however variations on the theme for the t-test. If you have a sample and wish to compare it with a known mean (e.g. national average) the single sample t-test is available. If both of your samples are not independent of each other and have some factor in common, i.e. geographical location or before/after treatment, the paired sample t-test can be applied. There are also two variations on the two sample t-test, the first uses samples that do not have equal variances and the second uses samples whose variances are equal.






 Data types that can be analysed with z-tests

• data points should be independent from each other
•  z-test is preferable when n is greater than 30.
•  the distributions should be normal if n is low, if however n>30 the distribution of the data does not have to be normal
•  the variances of the samples should be the same (F-test) 
•  all individuals must be selected at random from the population
•  all individuals must have equal chance of being selected
•  sample sizes should be as equal as possible but some differences are allowed

    Data types that can be analysed with t-tests

• Data sets should be independent from each other except in the case of the paired-sample t-test
•  where n<30 the t-tests should be used
•  the distributions should be normal for the equal and unequal variance t-test (K-S test or Shapiro-Wilke)
•  the variances of the samples should be the same (F-test) for the equal variance t-test
•  all individuals must be selected at random from the population
•  all individuals must have equal chance of being selected
•  sample sizes should be as equal as possible but some differences are allowed

Example of z-test problem: 

      A researcher reports that the average salary assistant professors is more than 42 000. A sample of 30 assistant professors has a mean salary of 43, 260. At a=0.05, test the claim that assistant professors earn more than 42 000 a year. The standard deviation of the population is 5230. 

Step 1. State the hypothesis and identify the claim.  

Ho : μ ≤ 42 000     and   H1: μ   >  42 000 (claim) 

Step 2. Find the Critical Value. Since a=0.05 and the test is a right-tailed test, the critical value is z= 1.65. 



Step 3. Compute the Test Value 

Step 4. Make the decision. Since the test value, 1.32 is less than the critical value 1.65, and is not in the critical region, the decision is "Do not reject the null hypothesis." 


Step 5. Summarize the results. There is not enough evidence to support the claim that assistant professors earn more on average than 24 000 a year.

  

The four possible outcomes and the summary statement for each situation. 

The P-value Method

The Traditional method - Using Rejection Regions (critical value approach) 

Steps in Solving Hypothesis Testing Problems (P-Value Method)

Step 1. State the hypothesis and identify the claim.
Step 2. Compute the test value.
Step 3. Find the P-Value.
Step 4. Make the decision.
Step 5. Summarize the results.


Decision Rule Based on P-value
To use a P-value to make a conclusion in a hypothesis test, compare the P-value with α
1. If P ≤ α, then reject H0.
2. If P > α, then fail to reject H0.



Example 1: A researcher wishes to test the claim that the average age of lifeguard in Ocean City is grater than 24 years. She selects a sample of 36 lifeguards and find the mean of the sample to 24.7 years, with a standard deviation of 2 years. Is there evidence to support the claim at a=0.05? Find the P-Value.

Solution:
Step 1. State the hypothesis and identify the claim.
Ho: µ  24        H1: µ  24

Step 2. Compute the test value.

Step 3. Find the P-value. Using the Table E. in appendix C, find the corresponding area under the normal distribution for z= 2.10. If is 0.4821. Subtract this value from 0.50 to find the right tail.
0.5-.4821 + 0.0179.

Hence the p-value is 0.0179

Step 4. Make the decision. Since the p-value is less than 0.05, the decision is to reject the null hypothesis.

Step 5. Summarize the Results. There is enough evidence to support the claim that the average age of lifeguards in Ocean City is greater than 24 years.




By: Tito Nuevacobita Jr. III-Gold