To test the first figure, let p be the proportion of email users who feel that spam has increased in their personal email. The first set of hypotheses that the student wants to test is then:

Ho: p = 0.37

Ha: p ≠ 0.37

Based upon the data collected by this student, a 95% confidence interval for p was found to be:

(0.25, 0.32).

Based on the collected data, a 95% confidence interval for p was found to be (0.08, 0.14).

For testing the second figure in the report, let p be the proportion of email users who feel that spam has increased in their work email. The second set of hypotheses that the students wants to test, is then:

Ho: p = 0.29

Ha: p ≠ 0.29

Based upon the data collected by this student, a 95% confidence interval for p was found to be:

(0.273, 0.304).

According to a study completed in 2006 by Pew Internet, 42% of all Americans had a broadband Internet connection at home. This same statistics student wanted to see if this percentage is different for students at his university.

Ho: p = 0.42

Ha: p ≠ 0.42

Based upon the data the student collected, a 95% confidence interval for p was found to be:

(0.439, 0.457).

According to the same Pew Internet study, 8% of those with broadband connections are using fixed wireless. let p be the proportion of broadband users who use fixed wireless, and consider the hypotheses:

Ho: p = 0.08

Ha: p ≠ 0.08

This document is linked from More about Hypothesis Testing.

]]>Ho: p = 0.087

Ha: p ≠ 0.087

Based on the collected data, a 95% confidence interval for p was found to be (0.08, 0.14).

The UCLA Internet Report (February 2003) estimated that roughly 60.5% of U.S. adults use the Internet at work for personal use. A follow-up study was conducted in order to explore whether that figure has changed since. Let p be the proportion of U.S. adults who use the Internet at work for personal use.

Ho: p = 0.605

Ha: p ≠ 0.605

Based on the collected data, the p-value of the test was found to be 0.001.

This document is linked from More about Hypothesis Testing.

]]>- In test I, a significance level of .05 was used, and the p-value was calculated to be 0.025.

- In test II, a significance level of .01 was used, and the p-value was calculated to be 0.025.

This document is linked from Proportions (Step 4 & Summary).

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In each of the following questions, choose the pair(s) of hypotheses for the population proportion (p) and the z statistic that match the figure.

This document is linked from Proportions (Step 3).

]]>Consider the following hypotheses, which we have seen before.

**Ho:**The average time full-time undergraduate college students study outside of class per week is 30 hours.**Ha:**The average time full-time undergraduate college students study outside of class per week is not 30 hours.

In a previous example, we had the following hypotheses:

**Ho:**The mean concentration in the shipment is the required 245 ppm.**Ha:**The mean concentration in the shipment is not the required 245 ppm.

From the results obtained, we rejected the null hypothesis and concluded with very little doubt that the mean concentration in the shipment is not the required 245 ppm.

In another example, we had the following hypotheses.

**Ho:**Performance on the SAT is not related to gender (males and females score the same).**Ha:**Performance on the SAT is related to gender – males score higher.

The data did not provide enough evidence for rejecting Ho. So there was not enough evidence that males score higher than females. The difference observed was not statistically significant.

This document is linked from Errors and Power.

]]>Let p be the proportion of all U.S. Internet-using households who have dial-up access. In a previous activity, we established that the appropriate hypotheses here are:

Ho: p = 0.75 and Ha: p < 0.75

Ann and Sam are both testing the hypothesis that 40% of plain M&M’s are orange, Ho: p = 0.40. Ann draws a sample of M&M’s and 45% of the M&M’s in her sample are orange. She calculates a test statistic of z = 1.25. Sam draws a sample of M&M’s and 50% of the M&M’s in his sample are orange. He calculates a test statistic of z = 1.

What can we conclude? Mark each statement as true or false.

This document is linked from Proportions (Step 2).

]]>**Scenario 1:** When shirts are made, there can occasionally be defects (such as improper stitching). But too many such defective shirts can be a sign of substandard manufacturing.

Suppose, in the past, your favorite department store has had only one defective shirt per 200 shirts (a prior defective rate of only .005). But you suspect that the store has recently switched to a substandard manufacturer. So you decide to test to see if their overall proportion of defective shirts today is higher.

Suppose that, in a random sample of 200 shirts from the store, you find that 27 of them are defective, for a sample proportion of defective shirts of .135. You want to test whether this is evidence that the store is “guilty” of substandard manufacturing, compared to their prior rate of defective shirts.

**Scenario 2:** It is a known medical fact that just slightly fewer females than males are born (although the reasons are not completely understood); the known “proper” baseline female birthrate is about 49% females.

In some cultures, male children are traditionally looked on more favorably than female children, and there is concern that the increasing availability of ultrasound may lead to pregnant mothers deciding to abort the fetus if it’s not the culturally “desired” gender. If this is happening, then the proportion of females in those nations would be significantly lower than the proper baseline rate.

To test whether the proportion of females born in India is lower than the proper baseline female birthrate, a study investigates a random sample of 6,500 births from hospital files in India, and finds 44.8% females born among the sample.

**Scenario 3:** A properly-balanced 6-sided game die should give a 1 in exactly 1/6 (16.7%) of all rolls. A casino wants to test its game die. If the die is not properly balanced one way or another, it could give either too many 1’s or too few 1’s, either of which could be bad.

The casino wants to use the proportion of 1’s to test whether the die is out of balance. So the casino test-rolls the die 60 times and gets a 1 in 9 of the rolls (15%).

This document is linked from Proportions (Introduction & Step 1).

]]>This document linked from Proportions (Introduction & Step 1).

]]>The same researchers also wanted to examine whether second-hand smoking (exposure to another person smoking) by pregnant women increases the risk of low birth weight (i.e., the proportion of babies born at a low birth weight among women who were second-hand smokers during their pregnancy is higher than the proportion in the general population). The researchers obtained a sample of 175 pregnant women who were second-hand smokers, followed them during their pregnancies, and found that 10.2% of the newborns had low birth weight. Based on these data, the p-value was found to be 0.119.

This document is linked from Steps in Hypothesis Testing.

]]>- Ho: The average number of miles driven per year is 12,000.
- Ha: The average number of miles driven per year is less than 12,000.

In a survey, 1,600 randomly selected drivers were asked the number of miles they drive yearly. Based upon the results, the p-value = 0.068.

Comment: Throughout this activity use a 0.05 (5%) significance level (cutoff).

This document is linked from Steps in Hypothesis Testing.

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