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Sleep habits of New Yorkers. New York is known as "the city that never sleeps". A random sample of 25 New Yorkers were asked how much sleep they get per night. Statistical summaries of these data are shown below. Do these data provide strong evidence that New Yorkers sleep more or less than 8 hours a night on average? \begin{tabular}{rrrrr} \hline \(\mathrm{n}\) & \(\bar{x}\) & \(\mathrm{~s}\) & \(\min\) & \(\max\) \\ \hline 25 & 7.73 & 0.77 & 6.17 & 9.78 \\ \hline \end{tabular} (a) Write the hypotheses in symbols and in words. (b) Check conditions, then calculate the test statistic, \(T,\) and the associated degrees of freedom. (c) Find and interpret the p-value in this context. Drawing a picture may be helpful. (d) What is the conclusion of the hypothesis test? (e) If you were to construct a \(95 \%\) confidence interval that corresponded to this hypothesis test, would you expect 8 hours to be in the interval?

Short Answer

Expert verified
There is not enough evidence to conclude that New Yorkers sleep differently from 8 hours per night. The p-value is 0.093, and 8 hours would likely be in the 95% confidence interval.

Step by step solution

01

Formulate Hypotheses

First, we need to set up the null and alternative hypotheses for the test. The null hypothesis \( H_0 \) is that New Yorkers sleep 8 hours per night on average: \( H_0 : \mu = 8 \). The alternative hypothesis \( H_a \) is that New Yorkers do not sleep 8 hours per night on average, meaning they either sleep more or less than 8 hours: \( H_a : \mu eq 8 \). In words, we are testing whether the average sleep of New Yorkers differs from 8 hours.
02

Check Conditions and Compute Test Statistic

To perform a t-test, we need to check if the sample size is large enough and if the sample is approximately normally distributed. With a sample size of 25, which is greater than 30, the t-distribution is appropriate. Next, calculate the test statistic using the formula: \[T = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{7.73 - 8}{0.77/\sqrt{25}} = \frac{-0.27}{0.154} \approx -1.75.\]The degrees of freedom \( df \) is \( n - 1 = 24 \).
03

Find and Interpret the P-Value

Using a t-distribution table or calculator with 24 degrees of freedom, find the p-value associated with the calculated test statistic \( T = -1.75 \). The p-value represents the probability of observing a test statistic as extreme as \( T \) under the null hypothesis. In this case, the p-value is approximately 0.093.
04

Draw a Conclusion From the Hypothesis Test

With a p-value of 0.093, we compare it to a common significance level of 0.05. Since 0.093 is greater than 0.05, we fail to reject the null hypothesis. This means there is not enough evidence to suggest that New Yorkers sleep more or less than 8 hours per night on average.
05

Consider the Confidence Interval

A 95% confidence interval for the mean sleep time would not include 8 hours if the result was significant at the 5% level. However, because we failed to reject the null hypothesis at that level, we would expect the interval to include 8 hours. Thus, the interval likely covers values both less and greater than 8.

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Key Concepts

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

T-distribution
The t-distribution is a type of probability distribution that is symmetric and bell-shaped, similar to the normal distribution but with heavier tails. It is used instead of the normal distribution when dealing with small sample sizes or when the population standard deviation is unknown. Typically, it provides a more accurate estimate when working with smaller datasets. In the context of hypothesis testing, it allows us to make inferences about a population mean based on a sample mean, particularly when the data is approximated from a small sample size.

In this exercise, a sample size of 25 was used. Because the sample size is under 30, the t-distribution is a fit choice for our statistical analysis. The degrees of freedom, often denoted as \( df \), is one less than the sample size (\( df = n - 1 \)). In this scenario, \( df \) is 24, accounting for our determination when using t-distribution tables or calculators to find critical values or p-values corresponding to the data.
P-value
A p-value is a critical concept in hypothesis testing that measures the probability of obtaining results at least as extreme as the observed results, assuming that the null hypothesis is true. It helps us determine the strength of the evidence against the null hypothesis. The smaller the p-value, the greater the evidence that we should reject the null hypothesis in favor of the alternative hypothesis.

In this specific exercise, the p-value obtained was approximately 0.093. This means there is a 9.3% likelihood that the observed sample mean (or something more extreme) could occur if the true average sleep time were 8 hours. Traditionally, a p-value is compared against a significance level, such as 0.05. Since 0.093 is larger than 0.05, it indicates that the observed data is not sufficiently unusual to reject the null hypothesis at the 5% significance level.
Confidence Interval
A confidence interval provides a range of values that is likely to contain the true population parameter with a certain level of confidence, typically 95%. It gives an estimate of where we might expect the true mean to lie based on our sample data. The wider this interval, the less precise the estimate. Conversely, a narrower interval indicates greater precision.

In the context of the exercise, if the calculation had resulted in the rejection of the null hypothesis, a corresponding 95% confidence interval would not include the value 8. Since the null hypothesis was not rejected, suggesting New Yorkers might indeed average around 8 hours of sleep, this 95% interval likely encompasses 8. Therefore, the confidence interval would include values less than 8, 8 itself, and values greater than 8, affirming that there is no evident deviation from the hypothesized average.
Null and Alternative Hypotheses
In hypothesis testing, we start by establishing a null hypothesis ( (H_0) ) and an alternative hypothesis ( (H_a) ). The null hypothesis represents a statement of no effect or no difference, being a default or standard position that indicates no change from a baseline or status quo. The alternative hypothesis is what we are interested in proving, representing the possibility of an effect or a difference.

For the problem about New Yorkers' sleep time, the null hypothesis is stated as \( H_0 : \mu = 8 \), which claims that the average sleep duration for New Yorkers is 8 hours. On the other hand, the alternative hypothesis is \( H_a : \mu eq 8 \), suggesting that New Yorkers’ average sleep is different from 8 hours. It's crucial to clearly delineate these hypotheses at the start of a hypothesis test to guide the statistical analysis accurately.

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Most popular questions from this chapter

4.36 True or false, Part I. Determine if the following statements are true or false, and explain your reasoning for statements you identify as false. (a) When comparing means of two samples where \(n_{1}=20\) and \(n_{2}=40,\) we can use the normal model for the difference in means since \(n_{2} \geq 30\). (b) As the degrees of freedom increases, the T distribution approaches normality. (c) We use a pooled standard error for calculating the standard error of the difference between means when sample sizes of groups are equal to each other.

Find the p-value. An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given set of hypotheses and \(T\) test statistic. Also determine if the null hypothesis would be rejected at \(\alpha=0.05\). (a) \(H_{A}: \mu>\mu_{0}, n=11, T=1.91\) (c) \(H_{A}: \mu \neq \mu_{0}, n=7, T=0.83\) (b) \(H_{A}: \mu<\mu_{0}, n=17, T=-3.45\) (d) \(H_{A}: \mu>\mu_{0}, n=28, T=2.13\)

Gaming and distracted eating, Part I. A group of researchers are interested in the possible effects of distracting stimuli during eating, such as an increase or decrease in the amount of food consumption. To test this hypothesis, they monitored food intake for a group of 44 patients who were randomized into two equal groups. The treatment group ate lunch while playing solitaire, and the control group ate lunch without any added distractions. Patients in the treatment group ate 52.1 grams of biscuits, with a standard deviation of 45.1 grams, and patients in the control group ate 27.1 grams of biscuits, with a standard deviation of 26.4 grams. Do these data provide convincing evidence that the average food intake (measured in amount of biscuits consumed) is different for the patients in the treatment group? Assume that conditions for inference are satisfied. \(^{41}\)

Diamonds, Part I. Prices of diamonds are determined by what is known as the 4 Cs: cut, clarity, color, and carat weight. The prices of diamonds go up as the carat weight increases, but the increase is not smooth. For example, the difference between the size of a 0.99 carat diamond and a 1 carat diamond is undetectable to the naked human eye, but the price of a 1 carat diamond tends to be much higher than the price of a 0.99 diamond. In this question we use two random samples of diamonds, 0.99 carats and 1 carat, each sample of size \(23,\) and compare the average prices of the diamonds. In order to be able to compare equivalent units, we first divide the price for each diamond by 100 times its weight in carats. That is, for a 0.99 carat diamond, we divide the price by 99. For a 1 carat diamond, we divide the price by \(100 .\) The distributions and some sample statistics are shown below. \(^{37}\) Conduct a hypothesis test to evaluate if there is a difference between the average standardized prices of 0.99 and 1 carat diamonds. Make sure to state your hypotheses \begin{tabular}{|l|} \hline \\ \hline \end{tabular} clearly, check relevant conditions, and interpret your results in context of the data.

Child care hours, Part I. The China Health and Nutrition Survey aims to examine the effects of the health, nutrition, and family planning policies and programs implemented by national and local governments. One of the variables collected on the survey is the number of hours parents spend taking care of children in their household under age 6 (feeding, bathing, dressing, holding, or watching them). In 2006,487 females and 312 males were surveyed for this question. On average, females reported spending 31 hours with a standard deviation of 31 hours, and males reported spending 16 hours with a standard deviation of 21 hours. Calculate a \(95 \%\) confidence interval for the difference between the average number of hours Chinese males and females spend taking care of their children under age 6 . Also comment on whether this interval suggests a significant difference between the two population parameters. You may assume that conditions for inference are satisfied. \(^{34}\)

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