Questions: A wedding website states that the average cost of a wedding is 29,065. One concerned bride hopes that the average is less than reported. To see if her hope is correct, she surveys 45 recently married couples and finds that the average cost of weddings in the sample was 27,360. Assuming that the population standard deviation is 6097, is there sufficient evidence to support the bride's hope at the 0.01 level of significance?

A wedding website states that the average cost of a wedding is 29,065. One concerned bride hopes that the average is less than reported. To see if her hope is correct, she surveys 45 recently married couples and finds that the average cost of weddings in the sample was 27,360. Assuming that the population standard deviation is 6097, is there sufficient evidence to support the bride's hope at the 0.01 level of significance?
Transcript text: A wedding website states that the average cost of a wedding is $\$ 29,065$. One concerned bride hopes that the average is less than reported. To see if her hope is correct, she surveys 45 recently married couples and finds that the average cost of weddings in the sample was $\$ 27,360$. Assuming that the population standard deviation is $\$ 6097$, is there sufficient evidence to support the bride's hope at the 0.01 level of significance? Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places.
failed

Solution

failed
failed

Solution Steps

Step 1: Calculate the Standard Error

To determine the standard error \( SE \) of the sample mean, we use the formula:

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{6097}{\sqrt{45}} \approx 908.89 \]

Step 2: Compute the Test Statistic

Next, we calculate the Z-test statistic using the formula:

\[ Z = \frac{\bar{x} - \mu_0}{SE} = \frac{27360 - 29065}{908.89} \approx -1.88 \]

Step 3: Determine the P-value

For a left-tailed test, we find the probability associated with the calculated Z-test statistic:

\[ P = T(z) \approx 0.03 \]

Final Answer

The test statistic is approximately \( Z \approx -1.88 \) and the P-value is approximately \( P \approx 0.03 \).

Thus, the final answer is:

\[ \boxed{Z = -1.88} \]

Was this solution helpful?
failed
Unhelpful
failed
Helpful

Questions asked by other users

failedAnswered
Identify this sentence as simple, compound, complex, compound/complex. I have to work hard at algebra; nevertheless, I enjoy the subject. - compound - compound/complex - simple - complex
failedAnswered
Aggregate expenditure, or the total amount of spending in the economy, equals A. household spending on durable goods plus business investment spending. B. household spending on durable goods plus household spending on nondurable goods. C. consumption spending plus planned investment spending plus government purchases plus net exports. D. total spending by households plus total spending by businesses.
failedAnswered
A disordered eating pattern that is characterized by an unhealthy preoccupation with the body being too thin or not muscular enough is called
failedAnswered
The SI unit for the electric field is: 1 m / s 1 J 1 N/C 1 N 1 C
failedAnswered
Look at this sentence: Harry abandoned the article on Oprah Winfrey halfway through because he thought it was too long. Select the definition of article that the sentence uses. article noun articles 1. a thing or object: She bought two articles of clothing. 2. a piece of writing in a magazine or newspaper: I just read an interesting article. 3. (grammar) the word a, an, or the: The article "an" is less common than "a" or "the."