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Chapter 10: Problem 7

A simple random sample of size \(n=40\) is drawn from a population. The sample mean is found to be \(108.5,\) and the sample standard deviation is found to be \(17.9 .\) Is the population mean greater than 100 at the \(\alpha=0.05\) level of significance?

Short Answer

Expert verified
Reject the null hypothesis. The population mean is greater than 100.

Step by step solution

01

- State the Hypotheses

The null hypothesis (H_0i) is that the population mean \(\text{\mu}\)i is equal to 100. The alternative hypothesis (H_1i) is that the population mean \(\text{\mu}\)i is greater than 100. Formally, H_0: \(\text{\mu} = 100\) and H_1: \(\text{\mu} > 100\).
02

- Determine the Test Statistic

Since the sample size is large ( n = 40i) and the population standard deviation is unknown, use the t-i statistic. The test statistic is calculated as \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \)i, where \(\bar{x} = 108.5\)i, \(\text{\mu}=100\)i, \(s = 17.9\)i, and \(n = 40\)i.
03

- Calculate the Test Statistic

Substitute the values into the formula: \[ t = \frac{108.5 - 100}{17.9 / \sqrt{40}} = \frac{8.5}{2.83} \approx 3.004 \]
04

- Determine the Critical Value

For a one-tailed test at the \(\alpha = 0.05\)i level of significance with \(df = n-1 = 39 \)i degrees of freedom, the critical value from the t-i distribution is \(t_{\alpha, df} \approx 1.685\)i.
05

- Compare the Test Statistic with the Critical Value

Since the calculated test statistic (3.004) is greater than the critical value (1.685), we reject the null hypothesis.
06

- Draw the Conclusion

There is sufficient evidence to conclude that the population mean is greater than 100 at the \(\alpha = 0.05\)i level of significance.

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Key Concepts

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

null hypothesis
In hypothesis testing, we start with what is known as the null hypothesis, often denoted as \(H_0\).
The null hypothesis is a statement we aim to test and either reject or fail to reject based on sample data.
For this example, the null hypothesis is \(H_0: \mu = 100\), which means that the population mean is 100.
It represents a position of no effect or no difference and serves as the default or starting assumption.
By setting up the null hypothesis, we can use statistical methods to determine whether there is sufficient evidence against it.
alternative hypothesis
The alternative hypothesis, denoted as \(H_1\) or \(H_a\), is what you want to prove.
It is a statement that indicates the presence of an effect or difference.
In our example, the alternative hypothesis is \(H_1: \mu > 100\), proposing that the population mean is greater than 100.
This hypothesis is considered if the evidence provided by the sample data suggests a significant deviation from the null hypothesis.
If we find strong enough evidence against \(H_0\), we will accept \(H_1\) as more likely.
t-statistic
To test the hypotheses, we calculate a test statistic. Here, we use the t-statistic due to the sample size and unknown population standard deviation.
The formula for the t-statistic is \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \] where \(\bar{x} = 108.5\) is the sample mean, \(\mu = 100\) is the population mean under \(H_0\), \(s = 17.9\) is the sample standard deviation, and \(n = 40\) is the sample size.
Plugging in the values, we get: \[ t = \frac{108.5 - 100}{17.9 / \sqrt{40}} = \frac{8.5}{2.83} \approx 3.004 \]
This t-statistic helps us understand how far our sample mean is from the null hypothesis mean in units of standard error.
critical value
The critical value is a point on the test's distribution that is compared to the test statistic to make a decision.
For a one-tailed test at the \(\backslash\backslashalpha = 0.05\) level of significance and 39 degrees of freedom (df = n-1), the critical value from the t-distribution table is approximately 1.685.
If our calculated t-statistic is greater than this critical value, we reject the null hypothesis.
In this example, since \`t = 3.004\` is greater than \1.685\, we reject \(H_0\) and conclude there is sufficient evidence that the population mean is greater than 100. This decision is based on the fact that the observed test statistic falls in the rejection region defined by our critical value.

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Most popular questions from this chapter

Sports announcers often say, "I wonder which player will show up to play today."This is the announcer's way of saying that the player's performance varies dramatically from game to game. Suppose that the standard deviation of the number of points scored by shooting guards in the NBA is 8.3. A random sample of 25 games played by Derrick Rose results in a sample standard deviation of 6.7 points. Assume that a normal probability plot indicates that the points scored are approximately normally distributed. Is Derrick Rose more consistent than other shooting guards in the NBA at the \(\alpha=0.10\) level of significance?

In \(1994,52 \%\) of parents of children in high school felt it was a serious problem that high school students were not being taught enough math and science. A recent survey found that 256 of 800 parents of children in high school felt it was a serious problem that high school students were not being taught enough math and science. Do parents feel differently today than they did in \(1994 ?\) (a) What does it mean to make a Type II error for this test? (b) If the researcher decides to test this hypothesis at the \(\alpha=0.05\) level of significance, determine the probability of making a Type II error if the true population proportion is 0.50. What is the power of the test? (c) Redo part (b) if the true proportion is 0.48 .

In Problems \(9-14,\) the null and alternative hypotheses are given. Determine whether the hypothesis test is lefi-tailed, right-tailed, or two-tailed. What parameter is being tested? \(H_{0}: \sigma=4.2\) \(H_{1}: \sigma \neq 4.2\)

One aspect of queuing theory is to consider waiting time in lines. A fast-food chain is trying to determine whether it should switch from having four cash registers with four separate lines to four cash registers with a single line. It has been determined that the mean wait-time in both lines is equal. However, the chain is uncertain about which line has less variability in wait time. From experience, the chain knows that the wait times in the four separate lines are normally distributed with \(\sigma=1.2\) minutes. In a study, the chain reconfigured five restaurants to have a single line and measured the wait times for 50 randomly selected customers. The sample standard deviation was determined to be \(s=0.84\) minute. Is the variability in wait time less for a single line than for multiple lines at the \(\alpha=0.05\) level of significance?

Statistical Significance versus Practical Significance \(A\) math teacher claims that she has developed a review course that increases the scores of students on the math portion of the SAT exam. Based on data from the College Board, SAT scores are normally distributed with \(\mu=515 .\) The teacher obtains a random sample of 1800 students, puts them through the review class, and finds that the mean SAT math score of the 1800 students is 519 with a standard deviation of 111 . (a) State the null and alternative hypotheses. (b) Test the hypothesis at the \(\alpha=0.10\) level of significance. Is a mean SAT math score of 519 significantly higher than \(515 ?\) (c) Do you think that a mean SAT math score of 519 versus 515 will affect the decision of a school admissions administrator? In other words, does the increase in the score have any practical significance? (d) Test the hypothesis at the \(\alpha=0.10\) level of significance with \(n=400\) students. Assume the same sample statistics. Is a sample mean of 519 significantly more than \(515 ?\) What do you conclude about the impact of large samples on the hypothesis test?
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