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Chapter 9: Problem 54

The recommended number of hours of sleep per night is 8 hours, but everybody "knows" that the average college student sleeps less than 7 hours. The number of hours slept last night by 10 randomly selected college students is listed here: $$\begin{array}{rrrrrrr}5.2 & 6.8 & 6.2 & 5.5 & 7.8 & 5.8 & 7.1 & 8.1 & 6.9 & 5.6\end{array}$$ Use a computer or calculator to complete the hypothesis test: \(H_{o}: \mu=7, H_{a}: \mu<7, \alpha=0.05\).

Short Answer

Expert verified
The result of the t-test will tell whether we should reject or fail to reject the null hypothesis. If the calculated t-statistic is less than the critical t-value, we reject the null hypothesis and conclude that the average college student sleeps less than 7 hours.

Step by step solution

01

Calculate the Mean

The first step is to calculate the mean or average of the data set. The mean is the sum of all values divided by the number of values. Add up the sleep hours of all 10 students, then divide by 10.
02

Calculate the Standard Deviation

Take the differences between each data point and the mean, square them, add them together, then divide by the total number of data points, and finally, take the square root of the result. This gives you the standard deviation of the data set, which measures the amount of variation or dispersion in the data set.
03

Perform the t-test

The t-test is used to determine whether there is a significant difference between the means of two groups. In this case, we only have one group, but we are comparing it to a hypothesized mean of 7 hours. The formula for the t-test statistic is: \(T = \frac{\bar{X} - \mu}{s/\sqrt{n}}\) where \( \bar{X} \) is the sample mean, \( \mu \) is the hypothesized mean, \( s \) is sample standard deviation, and \( n \) is the sample size. Perform this calculation.
04

Compare to Critical Value

Check the calculated t-statistic against the t-value from t distribution table for \( \alpha = 0.05 \) and \( df = n - 1 \) which corresponds to the given significance level and degrees of freedom. The null hypothesis is rejected if the calculated t-value is less than the critical t-value in the case of a left-tailed test, which is what we have here since \( H_{a}: \mu<7 \).

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a method used to decide if there is enough evidence to reject a hypothesis about a population parameter. In our exercise, we start with two hypotheses:

  • Null Hypothesis ( $H_0$): The average sleep duration is 7 hours ( $渭 = 7$).
  • Alternative Hypothesis ( $H_a$): The average sleep is less than 7 hours ( $渭 < 7$).
This test checks if the mean sleep duration is significantly different from 7 hours. We use a significance level ( $伪$) of 0.05, which means we have a 5% risk of incorrectly rejecting the null hypothesis. The aim is to collect evidence to show whether the actual mean is less than the hypothesized mean.
Sample Mean
The sample mean (\(\bar{X}\)) is crucial in the hypothesis test, as it represents the average of data points in our sample. To find it, we add all the sleep hours and divide by the number of students (10 in this case). This gives us the average sleep duration for the sample group.

The sample mean is the point estimate of the population mean, and it's used in calculating the t-test statistic. It helps to compare the actual data to what we would expect (7 hours) and see if there's enough evidence to support that the mean is less than 7 hours.
Standard Deviation
Standard deviation ( $s$) measures the amount of variation or spread in a set of values. In hypothesis testing, it's essential to understand the degree to which individual data points differ from the sample mean. Calculating it involves the following steps:

  • Find the difference between each sleep hour and the sample mean.
  • Square each difference.
  • Add all squared differences together.
  • Divide by the number of observations (for the sample standard deviation, use $n-1$).
  • Take the square root of that value.
This value helps assess how much the sleep hours vary and is used in the denominator of the t-test formula to standardize the test statistic.
Critical Value
In hypothesis testing, the critical value determines the cutoff point, beyond which we reject the null hypothesis. Given a significance level $伪 = 0.05$, and knowing this is a left-tailed test (since $H_a: 渭 < 7$), we find the critical t-value using a t-distribution table.

  • The degrees of freedom ( $df$) in our test is $n - 1$, where $n = 10$, so $df = 9$.
  • We look up the critical t-value corresponding to $伪 = 0.05$ in the t-distribution table.
We compare the calculated t-statistic with this critical value. If it's smaller than the critical t-value, it means there's evidence to reject the null hypothesis and accept that the average sleep might be less than 7 hours.

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Most popular questions from this chapter

In obtaining the sample size to estimate a proportion, the formula \(n=[z(\alpha / 2)]^{2} p q / E^{2}\) is used. If a reasonable estimate of \(p\) is not available, it is suggested that \(p=0.5\) be used because this will give the maximum value for \(n\). Calculate the value of \(p q=p(1-p)\) for \(p=0.1,0.2,0.3, \ldots, 0.8,0.9\) in order to obtain some idea about the behavior of the quantity \(p q\).

Calculate the test statistic \(z \star\) used in testing the following: a. \(H_{o}: p=0.70\) vs. \(H_{a}: p>0.70,\) with the sample \(n=300\) and \(x=224\) b. \(H_{o}: p=0.50\) vs. \(H_{a}: p<0.50,\) with the sample \(n=450\) and \(x=207\) c. \(H_{o}: p=0.35\) vs. \(H_{a}: p \neq 0.35,\) with the sample \(n=280\) and \(x=94\) d. \(H_{o}: p=0.90\) vs. \(H_{a}: p>0.90,\) with the sample \(n=550\) and \(x=508\)

The dry weight of a cork is another quality that does not affect the ability of the cork to seal a bottle, but it is a variable that is monitored regularly. The weights of the no. 9 natural corks \((24 \mathrm{mm}\) in diameter by \(45 \mathrm{mm}\) in length) have a normal distribution. Ten randomly selected corks were weighed to the nearest hundredth of a gram. Dry Weight (in grams) $$\begin{array}{rrrrrrrrr}3.26 & 3.58 & 3.07 & 3.09 & 3.16 & 3.02 & 3.64 & 3.61 & 3.02 & 2.79\end{array}$$ a. Does the preceding sample present sufficient reason to show that the standard deviation of the dry weights is different from 0.3275 gram at the 0.02 level of significance? A different random sample of 20 is taken from the same batch. Dry Weight (in grams) $$\begin{array}{llllllllll}\hline 3.53 & 3.77 & 3.49 & 3.24 & 3.00 & 3.41 & 3.33 & 3.51 & 3.02 & 3.46 \\\2.80 & 3.58 & 3.05 & 3.51 & 3.61 & 2.90 & 3.69 & 3.62 & 3.26 & 3.58 \\\\\hline\end{array}$$ b. Does the preceding sample present sufficient reason to show that the standard deviation of the dry weights is different from 0.3275 gram at the 0.02 level of significance? c. What effect did the two different sample standard deviations have on the calculated test statistic in parts a and b? What effect did they have on the \(p\) -value or critical value? Explain. d. What effect did the two different sample sizes have on the calculated test statistic in parts a and b? What effect did they have on the \(p\) -value or critical value? Explain.

Reliable Equipment has developed a machine, The Flipper, that will flip a coin with predictable results. They claim that a coin flipped by The Flipper will land heads up at least \(88 \%\) of the time. What conclusion would result in a hypothesis test, using \(\alpha=0.05,\) when 200 coins are flipped and the following results are achieved? a. 181 heads b. 172 heads c. 168 heads d. 153 heads

Variation in the life of a battery is expected, but too much variation would be of concern to the consumer, who would never know if the purchased battery might have a very short life. A random sample of 30 AA batteries of a particular brand produced a standard deviation of 350 hours. If a standard deviation of 288 hours (12 days) is considered acceptable, does this sample provide sufficient evidence that this brand of battery has greater variation than what is acceptable at the 0.05 level of significance? Assume battery life is normally distributed.
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