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A study \(^{20}\) conducted in June 2015 examines ownership of tablet computers by US adults. A random sample of 959 people were surveyed, and we are told that 197 of the 455 men own a tablet and 235 of the 504 women own a tablet. We want to test whether the survey results provide evidence of a difference in the proportion owning a tablet between men and women. (a) State the null and alternative hypotheses, and define the parameters. (b) Give the notation and value of the sample statistic. In the sample, which group has higher tablet ownership: men or women? (c) Use StatKey or other technology to find the pvalue.

Short Answer

Expert verified
The null hypothesis is \(P_M = P_W\) and the alternative hypothesis is \(P_M \neq P_W\). The difference in proportions is 0.033, showing a higher tablet ownership in women. The p-value to test the hypotheses must be calculated using statistical software.

Step by step solution

01

State the Null and Alternative Hypotheses and Define the Parameters

The null hypothesis (\(H_0\)) is that there is no difference in the proportions of tablet ownership between men and women. This is symbolized as \(P_M = P_W\), where \(P_M\) is the proportion of men who own a tablet and \(P_W\) is the proportion of women who own a tablet. The alternative hypothesis (\(H_A\)) is that there is a difference in the proportions of tablet ownership between men and women. This is symbolized as \(P_M \neq P_W\). The parameters in this case are \(P_M\) and \(P_W\).
02

Define the Sample Statistic and Identify the Group with Higher Tablet Ownership

The sample statistic is the difference in proportions between tablet owners who are men and those who are women. This is calculated as follows: Ratio of men owning tablets = 197 / 455 = 0.433. Ratio of women owning tablets = 235 / 504 = 0.466. Difference in proportions = 0.466 - 0.433 = 0.033. The group with higher tablet ownership in the sample is Women.
03

Calculation of p-value

To find the p-value, you need to conduct a test of proportions. Using appropriate statistical software or methods like the Z-test, you input the number of owners and the sample sizes for both groups. You then conduct a two-proportion Z-test or the Chi-Square test for independence (depending on the software). The value generated by this test will be your p-value, indicating the probability that the observed difference could have come about by chance under the null hypothesis. Please note that to carry out this step it is necessary to use statistical software: the exact way to enter values will depend on this software, and the specific number can't be given without running this test.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null and Alternative Hypotheses
Understanding the null and alternative hypotheses is crucial when conducting a difference in proportions test, like in the textbook exercise concerning tablet ownership by gender. The null hypothesis (\(H_0\)) serves as a starting point, implying no effect or no difference, which would mean both genders own tablets at the same rate (\(P_M = P_W\)).

The alternative hypothesis (\(H_A\)), on the other hand, suggests that there is a difference in the proportions (\(P_M eq P_W\)). This hypothesis attempts to demonstrate what we suspect might be true, that tablet ownership varies between men (\(P_M\)) and women (\(P_W\)). The objective of the test is to determine if we have enough evidence to reject the null hypothesis and accept the alternative hypothesis.
Sample Statistic
A sample statistic is a numerical summary about a sample that often estimates the unknown parameter of interest in the population. In the context of the tablet ownership study, the sample statistic is the difference in the proportion of tablet ownership between the sample of men and women.

This difference is calculated by subtracting the proportion of male tablet owners (\(197 / 455 = 0.433\)) from the proportion of female tablet owners (\(235 / 504 = 0.466\)), which gives a difference of 0.033. This difference indicates which group, in the sample, has higher ownership, and in this case, it is women, as the sample statistic is positive.
P-value Calculation
The p-value is a crucial component in hypothesis testing, representing the probability that the sample statistic would be as extreme as it is, or more so, if the null hypothesis were true. To calculate the p-value in our tablet ownership study, one would typically use a Z-test for two proportions, due to the normal distribution of sample proportions.

The calculation involves determining how many standard deviations away the observed sample statistic is from the expected value under the null hypothesis. Although the specific p-value is not provided due to the lack of statistical software output, it would be the key to deciding whether the difference in tablet ownership by gender observed in the sample could be due to random chance.
Statistical Significance
Statistical significance plays an important role in determining whether the findings from a study can be considered valid. If the calculated p-value is less than a pre-determined threshold (commonly \(\alpha = 0.05\)), the result is considered statistically significant.

This means the observed difference is unlikely to have occurred by chance, and we have enough evidence to reject the null hypothesis. In the context of the tablet ownership exercise, if the p-value were below this threshold, it would suggest a statistically significant difference in tablet ownership between men and women.
Tablet Ownership by Gender
The study seeks to determine whether there's an actual disparity between male and female tablet ownership. Initial results from the sample indicate a higher proportion of women owning tablets.

This particular exercise demonstrates the process of examining observed data to determine if it reflects a true trend in the broader population or occurs merely by chance. By applying tests for the difference in proportions, we understand better how gender may or may not influence technology ownership patterns, an insight that contributes to market research and sociological studies.

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Most popular questions from this chapter

A study \(^{54}\) shows that relationship status on Facebook matters to couples. The study included 58 college-age heterosexual couples who had been in a relationship for an average of 19 months. In 45 of the 58 couples, both partners reported being in a relationship on Facebook. In 31 of the 58 couples, both partners showed their dating partner in their Facebook profile picture. Men were somewhat more likely to include their partner in the picture than vice versa. However, the study states: "Females' indication that they are in a relationship was not as important to their male partners compared with how females felt about male partners indicating they are in a relationship." Using a population of college-age heterosexual couples who have been in a relationship for an average of 19 months: (a) A \(95 \%\) confidence interval for the proportion with both partners reporting being in a relationshipon Facebook is about 0.66 to 0.88 . What is the conclusion in a hypothesis test to see if the proportion is different from \(0.5 ?\) What significance level is being used? (b) A \(95 \%\) confidence interval for the proportion with both partners showing their dating partner in their Facebook profile picture is about 0.40 to \(0.66 .\) What is the conclusion in a hypothesis test to see if the proportion is different from \(0.5 ?\) What significance level is being used?

The same sample statistic is used to test a hypothesis, using different sample sizes. In each case, use StatKey or other technology to find the p-value and indicate whether the results are significant at a \(5 \%\) level. Which sample size provides the strongest evidence for the alternative hypothesis? Testing \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1}>p_{2}\) using \(\hat{p}_{1}-\hat{p}_{2}=0.8-0.7=0.10\) with each of the following sample sizes: (a) \(\hat{p}_{1}=24 / 30=0.8\) and \(\hat{p}_{2}=14 / 20=0.7\) (b) \(\hat{p}_{1}=240 / 300=0.8\) and \(\hat{p}_{2}=140 / 200=0.7\)

Do you think that students undergo physiological changes when in potentially stressful situations such as taking a quiz or exam? A sample of statistics students were interrupted in the middle of a quiz and asked to record their pulse rates (beats for a 1-minute period). Ten of the students had also measured their pulse rate while sitting in class listening to a lecture, and these values were matched with their quiz pulse rates. The data appear in Table 4.18 and are stored in QuizPulse10. Note that this is paired data since we have two values, a quiz and a lecture pulse rate, for each student in the sample. The question of interest is whether quiz pulse rates tend to be higher, on average, than lecture pulse rates. (Hint: Since this is paired data, we work with the differences in pulse rate for each student between quiz and lecture. If the differences are \(D=\) quiz pulse rate minus lecture pulse rate, the question of interest is whether \(\mu_{D}\) is greater than zero.) Table 4.18 Quiz and Lecture pulse rates for I0 students $$\begin{array}{lcccccccccc} \text { Student } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \text { Quiz } & 75 & 52 & 52 & 80 & 56 & 90 & 76 & 71 & 70 & 66 \\\ \text { Lecture } & 73 & 53 & 47 & 88 & 55 & 70 & 61 & 75 & 61 & 78 \\\\\hline\end{array}$$ (a) Define the parameter(s) of interest and state the null and alternative hypotheses. (b) Determine an appropriate statistic to measure and compute its value for the original sample. (c) Describe a method to generate randomization samples that is consistent with the null hypothesis and reflects the paired nature of the data. There are several viable methods. You might use shuffled index cards, a coin, or some other randomization procedure. (d) Carry out your procedure to generate one randomization sample and compute the statistic you chose in part (b) for this sample. (e) Is the statistic for your randomization sample more extreme (in the direction of the alternative) than the original sample?

A confidence interval for a sample is given, followed by several hypotheses to test using that sample. In each case, use the confidence interval to give a conclusion of the test (if possible) and also state the significance level you are using. A \(90 \%\) confidence interval for \(p_{1}-p_{2}: 0.07\) to 0.18 (a) \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1} \neq p_{2}\) (b) \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1}>p_{2}\) (c) \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1}

The same sample statistic is used to test a hypothesis, using different sample sizes. In each case, use StatKey or other technology to find the p-value and indicate whether the results are significant at a \(5 \%\) level. Which sample size provides the strongest evidence for the alternative hypothesis? Testing \(H_{0}: p=0.5\) vs \(H_{a}: p>0.5\) using \(\hat{p}=0.55\) with each of the following sample sizes: (a) \(\hat{p}=55 / 100=0.55\) (b) \(\hat{p}=275 / 500=0.55\) (c) \(\hat{p}=550 / 1000=0.55\)

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