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Chapter 9: Problem 55

It is claimed that the students at a certain university will score an average of 35 on a given test. Is the claim reasonable if a random sample of test scores from this university yields \(33,42,38,37,30,42 ?\) Complete a hypothesis test using \(\alpha=0.05 .\) Assume test results are normally distributed. a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

Short Answer

Expert verified
Based on the results from the p-value and classical approach, if we have enough evidence (p-value less than 0.05 or test statistic in the critical region), we could reject the null hypothesis and conclude that the claim is not reasonable. Otherwise, we do not reject the null hypothesis and conclude that the claim is reasonable.

Step by step solution

01

Setup Hypothesis and Compute Sample Statistics

Let鈥檚 set up our null hypothesis and alternate hypothesis. The null hypothesis will be \(H_0: \mu = 35\) and the alternate hypothesis will be \(H_1: \mu \neq 35\). Here, \(\mu\) represents the population mean test score. Then, calculate the sample mean and sample standard deviation from the given score list: \(33, 42, 38, 37, 30, 42\). The number of students, which is the sample size \(n\), is 6.
02

Calculate the Test Statistic

Compute the test statistic using the formula: \(Z = \frac{\overline{X} - \mu}{\frac{S}{\sqrt{n}}}\). Where \(\overline{X}\) is the sample mean, \(\mu\) is the population mean, \(S\) is the sample standard deviation, and \(n\) is the sample size.
03

Determine the Critical Value and P-value

According to the problem, we use \(\alpha = 0.05\) level of significance, and since this is a two-tailed test, the critical z-values are -1.96 and +1.96. Next, calculate the p-value. The p-value would be the probability that a z-score is more extreme than the test statistic. You can calculate p-value using the standard normal probability table or software.
04

Draw Conclusions Using the P-value Approach

If the p-value is less than the level of significance \(\alpha\), we reject the null hypothesis. Therefore, if the computed p-value is less than \(0.05\), we reject \(H_0: \mu = 35\). If the p-value is not less than \(0.05\), we don't have enough evidence to reject the null hypothesis.
05

Draw Conclusions Using the Classical Approach

Using the test statistic and critical value, if the calculated test statistic falls within the critical region, then we reject the null hypothesis. Therefore, if the test statistic we computed in Step 2 is less than -1.96 or greater than +1.96, we reject \(H_0: \mu = 35\). If it's not, we don't have enough evidence 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.

Sample Mean
The sample mean is a critical concept in hypothesis testing as it represents the average of a set of observations from a sample. It is denoted as \(\overline{X}\) and is calculated by adding up all the values in a sample and dividing the sum by the number of observations in the sample. In the context of the exercise, the sample mean was found using the test scores \(33, 42, 38, 37, 30, 42\).

To calculate the sample mean, we simply add these scores together and divide by the number of scores, which is 6. This gives us a sample mean \(\overline{X}\) that serves as the estimated average of the population mean \(\mu\) for the purpose of hypothesis testing. It is important to note that the sample mean is only an estimate of the true population mean, and it is subject to variability based on the particular sample chosen. This variability is one reason why hypothesis testing includes measures of uncertainty, such as the standard deviation and the p-value, to make inferences about the population.
Test Statistic Calculation
The test statistic calculation is a procedure that allows us to make a comparison between the sample data and the null hypothesis. The test statistic is essentially a standardized value that tells us how far the sample mean is from the hypothesized population mean, in units of the standard deviation.

In our exercise, we use the test statistic in the Z-test formula \(Z = \frac{\overline{X} - \mu}{\frac{S}{\sqrt{n}}}\) to determine how many standard deviations our sample mean \(\overline{X}\) is from the hypothesized mean \(\mu\). The numerator \(\overline{X} - \mu\) measures the difference while the denominator \(\frac{S}{\sqrt{n}}}\), known as the standard error, accounts for the standard deviation of the sample distribution and sample size. This calculation allows us to determine the p-value or compare the test statistic against critical values in hypothesis testing.
Standard Deviation
The standard deviation is a measure of the amount of variability or dispersion in a set of values. A low standard deviation means that the values tend to be close to the mean, whereas a high standard deviation indicates that the values are spread out over a wider range.

In hypothesis testing, the standard deviation of the sample \(S\) is used to quantify the spread of the data around the sample mean. It plays a significant role in calculating the test statistic and consequently in making statistical inferences about the population. In the context of the provided exercise, the standard deviation is used within the test statistic calculation to standardize the difference between the sample mean and the population mean, allowing us to determine whether this difference is statistically significant or attributable to random chance. Understanding standard deviation is fundamental to interpreting the results of a hypothesis test and assessing the reliability of the conclusions drawn from the sample data.

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Most popular questions from this chapter

Find the test statistic for the hypothesis test: a. \(H_{o}: \sigma^{2}=532\) versus \(H_{a}: \sigma^{2}>532\) using sample information \(n=18\) and \(s^{2}=785\) b. \(H_{o}: \sigma^{2}=52\) versus \(H_{a}: \sigma^{2} \neq 52\) using sample information \(n=41\) and \(s^{2}=78.2\)

a. Does it seem reasonable that the mean of the sampling distribution of observed values of \(p^{\prime}\) should be \(p,\) the true proportion? Explain. b. Explain why \(p^{\prime}\) is an unbiased estimator for the population \(p\).

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: \(H_{o}: \mu=7, H_{a}: \mu<7, \alpha=0.05\).

Construct \(90 \%\) confidence intervals for the binomial parameter \(p\) for each of the following pairs of values. Write your answers on the chart. $$\begin{array}{lllll} & \begin{array}{l}\text { Observed Proportion } \\\p^{\prime}=x / n\end{array} & \text { Sample Size } & \text { Lower limit } & \text { Upper limit } \\\\\hline \text { a. } & p^{\prime}=0.3 & n=30 & \\\\\text { b. } & p^{\prime}=0.7 & n=30 & \\\\\text { c. } & p^{\prime}=0.5 & n=10 & \\\\\text { d. } & p^{\prime}=0.5 & n=100 & \\\\\text { e. } & p^{\prime}=0.5 & n=1000 & \\\\\hline\end{array}$$ f. Explain the relationship between the answers to parts a and b. g. Explain the relationship among the answers to parts c-e.

A Cambridge Consumer Credit Index nationwide telephone survey of 1000 people found that most Americans are not easily swayed by the lure of reward points or rebates when deciding to use a credit card or pay by cash or check. The survey found that 2 out of 3 consumers do not even have credit cards offering reward points or rebates. Explain why you would be reluctant to use this information to construct a confidence interval estimating the true proportion of consumers who do not have credit cards offering reward points or rebates.
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