Math Problem Statement

A marketing researcher wants to estimate the mean amount spent per year​ ($) on a web site by membership member shoppers. Suppose a random sample of 100 membership member shoppers who recently made a purchase on the web site yielded a mean amount spent of ​$57 and a standard deviation of ​$52. Is there evidence that the population mean amount spent per year on the web site by membership member shoppers is different from ​$51​? ​(Use a 0.10 level of​ significance.) State the null and alternative hypotheses. Upper H 0​: mu ▼ less than or equals greater than less than not equals equals greater than or equals    enter your response here Upper H 1​: mu ▼ less than equals greater than or equals less than or equals not equals greater than    enter your response here ​(Type integers or decimals. Do not round. Do not include the​ $ symbol in your​ answer.) Part 2 Identify the critical​ value(s). The critical​ value(s) is/are    enter your response here. ​(Type an integer or a decimal. Round to two decimal places as needed. Use a comma to separate answers as​ needed.) Part 3 Determine the test statistic. The test​ statistic, t Subscript STAT​, is    enter your response here. ​(Type an integer or a decimal. Round to two decimal places as​ needed.) Part 4 State the conclusion. ▼ Do not reject Reject Upper H 0. There is ▼ sufficient insufficient evidence that the population mean spent by membership member customers is different from ​$51. Part 5 b. Determine the​ p-value and interpret its meaning. The​ p-value is    enter your response here. ​(Type an integer or a decimal. Round to three decimal places as​ needed.) Part 6 Interpret the meaning of the​ p-value. Select the correct answer below. A. The​ p-value is the probability of obtaining a sample mean that is equal to or more extreme than ​$6 away from ​$51 if the null hypothesis is true. B. The​ p-value is the probability of obtaining a sample mean that is equal to or more extreme than ​$6 above ​$51 if the null hypothesis is false. C. The​ p-value is the probability of obtaining a sample mean that is equal to or more extreme than ​$6 below ​$51 if the null hypothesis is false. D. The​ p-value is the probability of not rejecting the null hypothesis when it is false.

Solution

Part 1: State the Null and Alternative Hypotheses

We are testing whether the population mean amount spent by membership member shoppers is different from $51, so this suggests a two-tailed test.

  • Null hypothesis (H₀): The population mean is equal to $51.

    H0:μ=51H_0: \mu = 51

  • Alternative hypothesis (H₁): The population mean is not equal to $51.

    H1:μ51H_1: \mu \neq 51

Part 2: Identify the Critical Value(s)

Given that the level of significance (α\alpha) is 0.10, and it is a two-tailed test, we will divide the significance level equally between the two tails: α/2=0.05\alpha/2 = 0.05 for each tail.

The critical value for a two-tailed test with α=0.10\alpha = 0.10 can be found using a t-distribution table (since the sample size is 100, degrees of freedom df=n1=99df = n - 1 = 99).

Using the t-distribution for a two-tailed test at α=0.10\alpha = 0.10:

  • Critical value for a two-tailed test at α=0.10\alpha = 0.10 and 99 degrees of freedom is approximately:

    tcritical=±1.660t_{critical} = \pm 1.660

Part 3: Determine the Test Statistic

The test statistic for a t-test is calculated using the following formula:

tstat=xˉμ0snt_{stat} = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}

Where:

  • xˉ=57\bar{x} = 57 (sample mean),
  • μ0=51\mu_0 = 51 (hypothesized population mean),
  • s=52s = 52 (sample standard deviation),
  • n=100n = 100 (sample size).

Now, substituting the values:

tstat=575152100=65210=65.21.15t_{stat} = \frac{57 - 51}{\frac{52}{\sqrt{100}}} = \frac{6}{\frac{52}{10}} = \frac{6}{5.2} \approx 1.15

So, the test statistic is approximately:

tstat=1.15t_{stat} = 1.15

Part 4: State the Conclusion

  • The test statistic tstat=1.15t_{stat} = 1.15 is less than the critical value 1.6601.660.
  • Since 1.15<1.6601.15 < 1.660, we fail to reject the null hypothesis.

Thus, the conclusion is:

Do not reject H0.\text{Do not reject } H_0.

There is insufficient evidence that the population mean spent by membership member customers is different from $51.

Part 5: Determine the P-value

The p-value corresponds to the area under the t-distribution curve beyond the test statistic tstat=1.15t_{stat} = 1.15 in both tails (since it's a two-tailed test).

From a t-distribution table or using statistical software:

p-value0.254\text{p-value} \approx 0.254

Part 6: Interpret the Meaning of the P-value

The p-value is the probability of observing a sample mean as extreme as $6 away from $51 (either above or below) if the null hypothesis is true. The correct interpretation is:

Option A:

The p-value is the probability of obtaining a sample mean that is equal to or more extreme than $6 away from $51 if the null hypothesis is true.


Do you want more details or have any questions?

Here are 5 related questions to consider:

  1. What is the difference between one-tailed and two-tailed tests?
  2. How do you determine the degrees of freedom for a t-test?
  3. What does the critical value represent in hypothesis testing?
  4. How does sample size affect the outcome of a t-test?
  5. What is the significance level, and how do you choose it?

Tip: Always check whether you're performing a one-tailed or two-tailed test, as it directly affects the critical values and p-value interpretation.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Two-tailed t-test
Statistical Significance
P-value

Formulas

t_stat = (x̄ - μ₀) / (s / √n)
Critical value for t-distribution

Theorems

Central Limit Theorem
Student's t-distribution

Suitable Grade Level

Undergraduate (Statistics or Business Analytics)