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There are two major Internet providers in the Colorado Springs, Colorado, area, one called HTC and the other Mountain Communications. We want to investigate whether there is a difference in the proportion of times a customer is able to access the Internet. During a oneweek period, 500 calls were placed at random times throughout the day and night to HTC. A connection was made to the Internet on 450 occasions. A similar one-week study with Mountain Communications showed the Internet to be available on 352 of 400 trials. At the .01 significance level, is there a difference in the percent of time that access to the Internet is successful?

Short Answer

Expert verified
There is no significant difference in internet access success rates.

Step by step solution

01

State the Hypotheses

We start with formulating the null and alternative hypotheses. The null hypothesis (\( H_0 \)) states that there is no difference in the proportions, i.e., \( p_1 = p_2 \). The alternative hypothesis (\( H_a \)) states that the proportions are different, i.e., \( p_1 eq p_2 \).
02

Calculate Sample Proportions

Compute the sample proportions for both Internet providers. For HTC, the sample proportion is \( \hat{p_1} = \frac{450}{500} = 0.9 \). For Mountain Communications, the sample proportion is \( \hat{p_2} = \frac{352}{400} = 0.88 \).
03

Pool the Proportions and Find the Standard Error

The pooled sample proportion \( \hat{p} \) is calculated as follows: \( \hat{p} = \frac{450 + 352}{500 + 400} = \frac{802}{900} \approx 0.89 \). Then calculate the standard error for the difference between proportions: \( SE = \sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})} = \sqrt{0.89(1-0.89)(\frac{1}{500} + \frac{1}{400})} \approx 0.023 \).
04

Calculate the Z-Score

The Z-score is determined by the formula: \( Z = \frac{\hat{p_1} - \hat{p_2}}{SE} = \frac{0.9 - 0.88}{0.023} \approx 0.87 \).
05

Find Critical Z-value and Make a Decision

At a 0.01 significance level, we look at a Z-table to find the critical Z-value for a two-tailed test, which is approximately \( \pm 2.576 \). Since \( 0.87 \) (our calculated Z-score) is not beyond this range, we fail to reject the null hypothesis.

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Key Concepts

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

Proportion Comparison
When we are talking about proportion comparison, we aim to determine if there is a significant difference between two groups. In this context, we're analyzing whether two Internet providers differ in their success rates of establishing connections.

To do this, we first calculate the sample proportion for each provider. Proportion is basically the number of successful connections divided by the total number of attempts for each provider. For HTC, it's 450 successes out of 500 attempts, resulting in a proportion of 0.9. For Mountain Communications, it's 352 out of 400, equating to 0.88.

This sample proportion gives us a snapshot of each provider's performance, which we can use to compare against each other. The goal is to see if these proportions are statistically significantly different from each other, indicating that one provider is more reliable than the other in this aspect.
Z-Score Calculation
Calculating the Z-score is a crucial step in hypothesis testing, especially when comparing proportions. A Z-score measures the distance between the sample proportion difference and the null hypothesis of zero difference, in terms of standard error units.

Before calculation, we find the pooled proportion, representing a combined success rate from both samples. Here, it is approximately 0.89, found by adding the number of successes from both providers and dividing by the total trials.

Next, we compute the standard error (SE) for the difference in proportions, using the pooled proportion and the separate sample sizes. In this example, SE is approximately 0.023.

Finally, the Z-score itself is derived from subtracting one sample proportion from another, and dividing by this SE: \[ Z = \frac{0.9 - 0.88}{0.023} \approx 0.87. \]

The resulting Z-score informs us how far, and in what direction, the observed difference deviates from the hypothesized zero difference.
Null and Alternative Hypotheses
Hypothesis testing begins by forming two conflicting hypotheses: the null hypothesis (\( H_0 \)) and the alternative hypothesis (\( H_a \)). These guide our statistical inference.

The null hypothesis is the statement we initially assume to be true. It claims there's no change, effect, or difference. In this exercise, the null hypothesis suggests there's no difference in the proportion of successful Internet connections between the two providers, expressed as \( p_1 = p_2 \).

On the other hand, the alternative hypothesis challenges this assumption. It asserts there's a measurable difference between the two proportions, stated as \( p_1 eq p_2 \).

Deciding between these hypotheses usually involves calculating a test statistic (like a Z-score) and comparing it to a critical value (here found from a Z-table). If our test statistic falls outside a critical range, we have enough evidence to reject the null hypothesis, otherwise we "fail to reject" it, keeping the status quo. This method provides a structured way to assess claims with quantitative data.

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Most popular questions from this chapter

A recent study focused on the number of times men and women who live alone buy takeout dinners in a month. The information is summarized below. $$\begin{array}{|lcc|}\hline \text { Statistic } & \text { Men } & \text { Women } \\\\\hline \text { Mean } & 24.51 & 22.69 \\\\\text { Standard deviation } & 4.48 & 3.86 \\\\\text { Sample size } & 35 & 40 \\\\\hline\end{array}$$ At the .01 significance level, is there a difference in the mean number of times men and women order takeout dinners in a month? What is the \(p\) -value?

A cell phone company offers two plans to it subscribers. At the time new subscribers sign up, they are asked to provide some demographic information. The mean yearly income for a sample of 40 subscribers to Plan \(\mathrm{A}\) is \(\$ 57,000\) with a standard deviation of \(\$ 9,200 .\) This distribution is positively skewed; the actual coefficient of skewness is 2.11. For a sample of 30 subscribers to Plan B the mean income is \(\$ 61,000\) with a standard deviation of \(\$ 7,100 .\) The distribution of Plan B subscribers is also positively skewed, but not as severely. The coefficient of skewness is 1.54 . At the .05 significance level, is it reasonable to conclude the mean income of those selecting Plan \(\mathrm{B}\) is larger? What is the \(p\) -value? Do the coefficients of skewness affect the results of the hypothesis test? Why?

The null and alternate hypotheses are: $$\begin{array}{l}H_{0}: \pi_{1}=\pi_{2} \\\H_{1}: \pi_{1} \neq \pi_{2}\end{array}$$ A sample of 200 observations from the first population indicated that \(X_{1}\) is \(170 .\) A sample of 150 observations from the second population revealed \(X_{2}\) to be \(110 .\) Use the .05 significance level to test the hypothesis. a. State the decision rule. b. Compute the pooled proportion. c. Compute the value of the test statistic. d. What is your decision regarding the null hypothesis?

The owner of Bun 'N' Run Hamburger wishes to compare the sales per day at two locations. The mean number sold for 10 randomly selected days at the Northside site was 83.55, and the standard deviation was \(10.50 .\) For a random sample of 12 days at the Southside location, the mean number sold was 78.80 and the standard deviation was \(14.25 .\) At the .05 significance level, is there a difference in the mean number of hamburgers sold at the two locations? What is the \(p\) -value?

A recent study compared the time spent together by single- and dual-earner couples. According to the records kept by the wives during the study, the mean amount of time spent together watching television among the single-earner couples was 61 minutes per day, with a standard deviation of 15.5 minutes. For the dual-earner couples, the mean number of minutes spent watching television was 48.4 minutes, with a standard deviation of 18.1 minutes. At the .01 significance level, can we conclude that the single-earner couples on average spend more time watching television together? There were 15 single-earner and 12 dual-earner couples studied.

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