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91Ó°ÊÓ

A mail-order company claims that at least \(60 \%\) of all orders are mailed within 48 hours. From time to time the quality control department at the company checks if this promise is fulfilled. Recently the quality control department at this company took a sample of 400 orders and found that 208 of them were mailed within 48 hours of the placement of the orders. a. Testing at the \(1 \%\) significance level, can you conclude that the company's claim is true? b. What will your decision be in part a if the probability of making a Type I error is zero? Explain. c. Make the test of part a using the \(p\) -value approach and \(\alpha=.01\).

Short Answer

Expert verified
a. At the 1% significance level, company's claim is false and it mails less than 60% of orders within 48 hours. b. Even if the probability of making a Type I error is zero, the decision would remain the same. c. Using the p-value approach, we reject the null hypothesis since the p-value is less than 0.01.

Step by step solution

01

Formulate the Hypothesis

H0: p = 0.60 (Null hypothesis - the company's claim is true)\nHa: p < 0.60 (Alternative hypothesis - the company mails less than 60% of orders within 48 hours)
02

Compute Test Statistic

The test statistic is computed using the formula z = (p_hat - p) / sqrt((p*(1-p))/n), with p_hat being the sample proportion (208/400 = 0.52), p is the population proportion (0.60) and n is the size of the sample (400). This results in z approximately equal to -5.77.
03

Calculate Critical Value and Make Decision for Part a

The critical z value for a one-tail test at the 1% significance level is approximately -2.33. Since the computed z score of -5.77 is less than the critical value of -2.33, we reject the null hypothesis. Thus, at the 1% significance level, there is evidence to suggest that the company mails less than 60% of orders within 48 hours.
04

Make Decision for Part b

If the probability of making a Type I error is zero, this means that we are 100% sure that our decision to reject the null hypothesis is correct. Therefore, the decision will remain the same as in Part a: we reject the null hypothesis.
05

Calculate P-value and Make Decision for Part c

The P-value is the probability of obtaining a test statistic as extreme as our result, given that the null hypothesis is true. Finding the p-value from the z-table for the calculated value of z = -5.77, we see it is less than 0.0001, which is less than our given level of significance (alpha = 0.01). Therefore, we reject the null hypothesis using the p-value approach.

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Key Concepts

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

Null Hypothesis
The null hypothesis is a cornerstone of statistical hypothesis testing. It is a statement claiming no effect or no difference in the context of the experiment or observation. In this case, the null hypothesis is that at least 60% of the company's orders are mailed within 48 hours, as claimed by the company. Mathematically, this can be stated as \( H_0: p = 0.60 \).

The null hypothesis is essentially the default assumption or the "status quo". In many scenarios, it is the hypothesis that researchers strive to test against, hoping to find evidence to refute it. When we perform hypothesis testing, we either find sufficient evidence to reject the null hypothesis or fail to reject it. Failing to reject it doesn’t necessarily mean it is true; rather, it implies that there isn't enough evidence against it given the sample data.

In hypothesis testing, rejecting the null hypothesis suggests that there might be something more significant happening, prompting further investigation or changes.
Alternative Hypothesis
The alternative hypothesis stands in contrast to the null hypothesis. It is what a researcher wants to prove or what they hope might be true instead of the null hypothesis. In this example, the alternative hypothesis is \( H_a: p < 0.60 \), implying that less than 60% of the orders are mailed within 48 hours.

This hypothesis suggests a deviation from the company's claim and is essentially trying to find a reason for action or change. It represents the new theory or idea, indicating that current conditions might not be as expected, which could necessitate intervention.

Testing involves looking for evidence to support the alternative hypothesis enough to reject the null hypothesis. Keep in mind, rejecting the null hypothesis in favor of the alternative implies there’s statistically sufficient evidence to consider the possibility of a difference or effect.
Significance Level
The significance level, often denoted by \( \alpha \), is a critical concept in hypothesis testing indicating how confident we need to be in our test results to reject the null hypothesis. It sets the threshold for determining whether an observed effect is statistically significant.

In this problem, the significance level is set at 1% (\( \alpha = 0.01 \)). This means that there is only a 1% risk of concluding that a difference exists when in fact, it does not (Type I error).

A low significance level, such as 0.01, implies a high standard for evidence required to reject the null hypothesis. It ensures that the findings are not due to random chance but instead signify a more consistent and reliable difference or effect. The choice of significance level reflects how the risk of incorrectly rejecting the null hypothesis is balanced against the need for scientific context and evidence.
P-value
The p-value helps in making decisions about the null hypothesis by quantifying the probability of observing data as extreme as what was collected assuming that the null hypothesis is true. It assists in determining the strength of the evidence against the null hypothesis.

For the given test, a p-value of less than 0.0001 was obtained with the test statistic. This is significantly lower than the chosen significance level of 1% (\( \alpha = 0.01 \)).

A very low p-value (like in this case) suggests that the observed data is highly unlikely under the null hypothesis. Thus, we reject the null hypothesis, reinforcing the conclusion that the percentage of orders mailed within 48 hours is indeed less than the claimed 60%.

Interpreting p-values:
  • If the p-value is less than \( \alpha \), we reject the null hypothesis.
  • If the p-value is greater than \( \alpha \), we do not reject the null hypothesis.
  • A small p-value indicates strong evidence against the null hypothesis.

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Most popular questions from this chapter

A company claims that the mean net weight of the contents of its All Taste cereal boxes is at least 18 ounces. Suppose you want to test whether or not the claim of the company is true. Explain briefly how you would conduct this test using a large sample. Assume that \(\sigma=.25\) ounce.

Explain how the tails of a test depend on the sign in the alternative hypothesis. Describe the signs in the null and alternative hypotheses for a two-tailed, a left-tailed, and a right-tailed test, respectively.

A study claims that all adults spend an average of 14 hours or more on chores during a weekend. A researcher wanted to check if this claim is true. A random sample of 200 adults taken by this researcher showed that these adults spend an average of \(14.65\) hours on chores during a weekend. The population standard deviation is known to be \(3.0\) hours. a. Find the \(p\) -value for the hypothesis test with the alternative hypothesis that all adults spend more than 14 hours on chores during a weekend. Will you reject the null hypothesis at \(\alpha=.01\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.01\).

The past records of a supermarket show that its customers spend an average of \(\$ 95\) per visit at this store. Recently the management of the store initiated a promotional campaign according to which each customer receives points based on the total money spent at the store, and these points can be used to buy products at the store. The management expects that as a result of this campaign, the customers should be encouraged to spend more money at the store. To check whether this is true, the manager of the store took a sample of 14 customers who visited the store. The following data give the money (in dollars) spent by these customers at this supermarket during their visits. \(\begin{array}{rrrrrrr}109.15 & 136.01 & 107.02 & 116.15 & 101.53 & 109.29 & 110.79 \\ 94.83 & 100.91 & 97.94 & 104.30 & 83.54 & 67.59 & 120.44\end{array}\) Assume that the money spent by all customers at this supermarket has a normal distribution. Using the \(5 \%\) significance level, can you conclude that the mean amount of money spent by all customers at this supermarket after the campaign was started is more than \(\$ 95 ?\) (Hint: First calculate the sample mean and the sample standard deviation for these data using the formulas learned in Sections \(3.1 .1\) and \(3.2 .2\) of Chapter \(3 .\) Then make the test of hypothesis about \(\mu .\) )

Consider the null hypothesis \(H_{0}: \mu=100\). Suppose that a random sample of 35 observations is taken from this population to perform this test. Using a significance level of \(.01\), show the rejection and nonrejection regions and find the critical value(s) of \(t\) when the alternative hypothesis is as follows. a. \(H_{1}: \mu \neq 100\) b. \(H_{1}: \mu>100\) c. \(H_{1}: \mu<100\)

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