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Do You Own a Smartphone? A study \(^{19}\) conducted in July 2015 examines smartphone ownership by US adults. A random sample of 2001 people were surveyed, and the study shows that 688 of the 989 men own a smartphone and 671 of the 1012 women own a smartphone. We want to test whether the survey results provide evidence of a difference in the proportion owning a smartphone 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 smartphone ownership: men or women? (c) Use StatKey or other technology to find the pvalue.

Short Answer

Expert verified
The null hypothesis (\(H_0\)) is that there is no difference between the proportion of men and women owning a smartphone. The alternative hypothesis (\(H_a\)) is that there is a difference. The sample proportions are approximately 0.696 for men and 0.663 for women, with men having a higher rate of ownership. Using a statistical software tool, the P-value can be computed providing evidence against the null hypothesis.

Step by step solution

01

Formulate the Hypotheses

The null hypothesis (\(H_0\)) refers to the situation where there is no difference in the proportion of men and women owning a smartphone. The alternative hypothesis (\(H_a\)) refers to there being a difference in the proportions. So, \(H_0: p_m = p_w\) and \(H_a: p_m ≠ p_w\) where \(p_m\) is the proportion of men that own a smartphone and \(p_w\) is the proportion of women that own a smartphone.
02

Calculate the Sample Statistics

The sample proportions for men and women owning a smartphone are computed. For men, it’s 688 out of 989, giving a sample statistic of approximately 0.696. For women, it’s 671 out of 1012, giving a sample statistic of approximately 0.663. The notation to represent these would be \(\hat{p}_m\) for men and \(\hat{p}_w\) for women. Comparing these, it can be observed that men have a higher smartphone ownership than women in the sample.
03

Calculate the P-value

To calculate the P-value, a statistical software or calculator would be needed. The function in the software would require the sample size and the number of successful outcomes for each group. In this case, the sample sizes are 989 and 1012 for men and women respectively, and the successful outcomes (owning a smartphone) are 688 and 671. After inputting the appropriate values, the P-value obtained gives the strength of evidence against the null hypothesis.

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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 to hypothesis testing in statistics. In the context of our exercise, the null hypothesis (\(H_0\)) is that there is no significant difference in the proportions of men and women owning smartphones. Mathematically, this is expressed as \(p_m = p_w\), where \(p_m\) and \(p_w\) are the population proportions of smartphone ownership among men and women, respectively. The alternative hypothesis (\(H_a\)), on the other hand, posits that there is a difference in these proportions (\(p_m eq p_w\)).

It might seem counterintuitive, but in hypothesis testing, we actually assume the null hypothesis is true and then look for evidence to disprove it. If sufficient evidence is found (usually through calculating the p-value), we might reject the null hypothesis in favor of the alternative. It's a bit like a courtroom where the null hypothesis is 'innocent until proven guilty'.
Sample Proportion
The sample proportion is a statistic that estimates the equivalent population parameter. In our example, the sample proportion is the number of individuals with a certain characteristic (owning a smartphone) divided by the total number of individuals in the sample. To denote the sample proportions, we use \(\hat{p}\).

Here's how you would calculate it: For men, \(\hat{p}_m = \frac{688}{989}\), and for women, \(\hat{p}_w = \frac{671}{1012}\). When working with sample proportions, it's important to understand that these are only estimations of the true population proportions. The sample values can help us infer about the population, but they themselves are prone to varying from one sample to another—a concept known as sampling variability. Furthermore, the accuracy of the sample proportion as an estimator depends on the sample size and how representative the sample is of the population.
P-value Calculation
The p-value is a crucial component in hypothesis testing, as it helps determine the strength of the evidence against the null hypothesis. It's calculated by assessing the probability of obtaining sample results as extreme as the ones observed if the null hypothesis were true.

In our smartphone ownership example, to find the p-value, we would need to use statistical software or a calculator. The software would take into account both the observed sample proportions and the sizes of the samples. Generally, if this calculated p-value is less than a predetermined significance level (often \(\alpha = 0.05\)), we reject the null hypothesis. A low p-value indicates that the observed data would be highly unlikely if the null hypothesis were true, suggesting that the alternative hypothesis may be a more plausible explanation. Conversely, a high p-value implies that the observed data are consistent with a true null hypothesis, so we wouldn't reject \(H_0\) in this scenario.

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Most popular questions from this chapter

Approval Rating for Congress In a Gallup poll \(^{51}\) conducted in December 2015 , a random sample of \(n=824\) American adults were asked "Do you approve or disapprove of the way Congress is handling its job?" The proportion who said they approve is \(\hat{p}=0.13,\) and a \(95 \%\) confidence interval for Congressional job approval is 0.107 to 0.153 . If we use a 5\% significance level, what is the conclusion if we are: (a) Testing to see if there is evidence that the job approval rating is different than \(14 \%\). (This happens to be the average sample approval rating from the six months prior to this poll.) (b) Testing to see if there is evidence that the job approval rating is different than \(9 \%\). (This happens to be the lowest sample Congressional approval rating Gallup ever recorded through the time of the poll.)

In this exercise, we see that it is possible to use counts instead of proportions in testing a categorical variable. Data 4.7 describes an experiment to investigate the effectiveness of the two drugs desipramine and lithium in the treatment of cocaine addiction. The results of the study are summarized in Table 4.14 on page \(323 .\) The comparison of lithium to the placebo is the subject of Example 4.34 . In this exercise, we test the success of desipramine against a placebo using a different statistic than that used in Example 4.34. Let \(p_{d}\) and \(p_{c}\) be the proportion of patients who relapse in the desipramine group and the control group, respectively. We are testing whether desipramine has a lower relapse rate then a placebo. (a) What are the null and alternative hypotheses? (b) From Table 4.14 we see that 20 of the 24 placebo patients relapsed, while 10 of the 24 desipramine patients relapsed. The observed difference in relapses for our sample is $$\begin{aligned}D &=\text { desipramine relapses }-\text { placebo relapses } \\\&=10-20=-10\end{aligned}$$ If we use this difference in number of relapses as our sample statistic, where will the randomization distribution be centered? Why? (c) If the null hypothesis is true (and desipramine has no effect beyond a placebo), we imagine that the 48 patients have the same relapse behavior regardless of which group they are in. We create the randomization distribution by simulating lots of random assignments of patients to the two groups and computing the difference in number of desipramine minus placebo relapses for each assignment. Describe how you could use index cards to create one simulated sample. How many cards do you need? What will you put on them? What will you do with them?

Does the airline you choose affect when you'll arrive at your destination? The dataset DecemberFlights contains the difference between actual and scheduled arrival time from 1000 randomly sampled December flights for two of the major North American airlines, Delta Air Lines and United Air Lines. A negative difference indicates a flight arrived early. We are interested in testing whether the average difference between actual and scheduled arrival time is different between the two airlines. (a) Define any relevant parameter(s) and state the null and alternative hypotheses. (b) Find the sample mean of each group, and calculate the difference in sample means. (c) Use StatKey or other technology to create a randomization distribution and find the p-value. (d) At a significance level of \(\alpha=0.01\), what is the conclusion of the test? Interpret the conclusion in context.

Interpreting a P-value In each case, indicate whether the statement is a proper interpretation of what a p-value measures. (a) The probability the null hypothesis \(H_{0}\) is true. (b) The probability that the alternative hypothesis \(H_{a}\) is true. (c) The probability of seeing data as extreme as the sample, when the null hypothesis \(H_{0}\) is true. (d) The probability of making a Type I error if the null hypothesis \(H_{0}\) is true. (e) The probability of making a Type II error if the alternative hypothesis \(H_{a}\) is true.

We are conducting many hypothesis tests to test a claim. In every case, assume that the null hypothesis is true. Approximately how many of the tests will incorrectly find significance? 800 tests using a significance level of \(5 \%\).

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