Hypothesis Testing for Proportions

I started these but am not sure how to finish them, any help would be appreciated.(Headbang)

1. Use the method specified to perform the hypothesis test for the population mean m. A local politician, running for reelection, claims that the mean prison time for car thieves is less than the required 6 years. A sample of 80 convicted car thieves was randomly selected. The mean of the sample was found to be 5.5 years. The population standard deviation is known to be 1.25 years. At a = 0.05, test the politician’s claim

a. Use the critical value z0 method from the normal distribution.
1. H0 : µ ≥ 6
Ha : µ < 6
2. a = 0.05
3. Test statistics: left-tailed
4. P-value or critical z0 or t0. -1.645
5. Rejection Region: z < -1.645

6. Decision: 6-5.5 = 0.5/1.25√80 =-3.58
7. Interpretation: There is not enough evidence at the 5% level of significance to support the local politician’s claim that the mean prison time for car thieves is less than 6.

b. Use the P-value method.


1. H0 :
Ha :
2. a =
3. Test statistics:
4. P-value or critical z0 or t0.
5. Rejection Region:

6. Decision:
7. Interpretation:


Question 2

Hypothesis Testing for Proportions.
The engineering school at a Northern university claims that 20% of its graduates are women. In a graduating class of 210 students, 58 were women. Does this suggest that the school is believable? Use a level of significance of a = 0.05 to test the school’s claim. Use an alternative hypothesis of p ¹ 0.2 (Round phat to 4 decimal places.)


1. H0 : p≥0.2
Ha : p<0.2
2. a = 0.05
3. Test statistics: 2.76 0.42-0.2/(0.2)(0.8)/210=0.95
4. P-value or critical z0 or t0.
5. Rejection Region:

6. Decision:
7. Interpretation:

Quote:

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7. Interpretation: There is not enough evidence at the 5% level of significance to support the local politician’s claim that the mean prison time for car thieves is less than 6.

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If you get a test statistic that is 3.5 standard deviations away from the population mean, you REJECT the null hypothesis. Remember we are testing the null hypothesis. If your test statistic fell within the non-reject region, we would conclude that there is not enough evidence to warrant rejection of the null hypothesis. Similarly, if our test statistic falls OUTSIDE of the non-reject region, we could concluse that there is enough evidence to warrannt rejection of the null hypothesis. Therefore we can say that there is evidence to suggest that the average time is probably less than the 6 years the politician is complaining about.

Hypothesis testing can be confusing, especially if you do not have an instructor who explains the idea behind it. Here, we are dealing with someone who is claiming that for the ENTIRE population of prisoners, that the time spent in jail is less than 6. If we assume that the actual population of car-thieves jail time IS 6 years, and we take a sample of 80 prisoners, find the average of their jail terms is 5.5 - we see, from computing the standard deviation for the AVERAGES of jail times, that it is off from the population mean by 3.5 standard deviations. This tells us one of two things, either our null hypothesis (that the average jail time for the POPULATION is 6 years) is false, and warrants rejection - OR that our null hypothesis is actually true, and we have managed to select a group of inmates whose average jail terms are indeed 3.5 standard deviations away from the actual mean of 6 years. As calculated above, the chances of that happening are 0.0002. Not very confidence inspiring.

For part B, using the P-Value, you simply compute the test statistic, and compare it against a P-Value of 0.05 (5%). The test statistic is obviously less than 0.05 so we can reject the null hypothesis.

See if you can tackle Question 2 now (remembering to adjust your hypothesis test to adjust for the statistics you are comparing - proportion instead of mean).