Question

A wedding website states that the average cost of a wedding is $25,809. One concerned bride hopes that the average is less than reported. To see if her hope is correct, she surveys 46 recently married couples and finds that the average cost of weddings in the sample was $24,638. Assuming that the population standard deviation is $5531, is there sufficient evidence to support the bride’s hope at the 0.10 level of significance?
Step 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below.
H0Ha: μ=25,809: μ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯25,809
Step 2 of 3 : Find the test statistic
Step 3 of 3 : What is the conclusion?

Answer 1

Answer:

1)

Null hypothesis            H₀ : μ = 25,809

Alternative hypothesis H₁ : μ < 25,809

2) Test Statistic = -1.44

3) Conclusion:

The result is significant, there is sufficient evidence to support the bride’s hope at the 0.10 level of significance.

Step-by-step explanation:

Given the data in the question;

Sample mean x" = 24,638

sample size n = 46

standard deviation σ = 5531

level of significance ∝ = 0.10

NULL and ALTERNATIVE HYPOTHESIS

Null hypothesis            H₀ : μ = 25,809

Alternative hypothesis H₁ : μ < 25,809

TEST STATISTICS

Z = (x"-μ) / σ√n

we substitute

Z = (24,638 - 25,809) / (5531/√46)

Z = -1171 / 815.5

Z = -1.44

Test Statistic = -1.44

Now, from normal z-table;

P-value = P( Z < -1.44 ) = 0.0749

P-value = 0.0749

Since P-value ( 0.0749 ) is less than level of significance ( 0.10 ), we reject H₀.

Conclusion:

The result is significant, there is sufficient evidence to support the bride’s hope at the 0.10 level of significance.