The manager of a bakery states in a report that the average value of customer transactions is greater than $13.65. A random sample of 38 transactions was taken, yielding an average of $14.92 with a standard deviation of $5.51. (a) Which hypotheses should be used to show the manager's claim is correct? H0: p = 14.92 vs. Ha: p > 14.92 H0: p = 13.65 vs. Ha: p > 13.65 H0: μ = 13.65 vs. Ha: μ > 13.65 H0: μ = 14.92 vs. Ha: μ > 14.92 (b) Find the test statistic: (Use 4 decimals.) (c) What is the p-value? (Use 4 decimals.) (d) What is the conclusion of the hypothesis test, for α = 0.05? --- the value $13.65. There is --- evidence the mean transaction value is greater than $13.65. (e) Regardless of your conclusion above, what is the meaning of a Type II error in this context? you conclude the average transaction value is greater than $13.65, in reality the average transaction value is greater than $13.65. You conclude the average transaction value is equal to $13.65, in reality the average transaction value is equal to $13.65. You conclude the average transaction value is greater than $13.65, in reality the average transaction value is equal to $13.65. You conclude the average transaction value is equal to $13.65, in reality the average transaction value is greater than $13.65.

A) Null Hypothesis; H0: μ = 13.65

Alternative hypothesis; Ha: μ > 13.65

B) t = 1.42

C) The p-value is greater than the significance level of 0.05. Thus, we fail to reject the null hypothesis and conclude that there is insufficient evidence to support the claim that the mean transaction value is greater than $13.65.

D) type II error will be to conclude the average transaction value is equal to $13.65, in reality the average transaction value is greater than $13.65.

Step-by-step explanation:

We are given;

Population mean; μ = $13.65

Sample mean; x¯ = $14.92

Sample standard deviation; s = $5.51

Sample size; n = 38

DF = n - 1 = 38 - 1 = 37

Significance level; α = 0.05

A) The hypotheses is defined below;

Null Hypothesis; H0: μ = 13.65

Alternative hypothesis; Ha: μ > 13.65

B) Since we are given sample standard deviation, we will use formula for t-value. Formula for the test statistic is;

t = (x¯ - μ)/(s/√n)

t = (14.92 - 13.65)/(5.51/√38)

t = 1.42

C) From online p-value from t-score calculator attached, using; t = 1.42, α = 0.05, DF = 37, one tail, we have;

P-value = 0.08199

D) The p-value is greater than the significance level of 0.05. Thus, we fail to reject the null hypothesis and conclude that there is insufficient evidence to support the claim that the mean transaction value is greater than $13.65.

E) A type II error simply means accepting a null hypothesis that is not true.

Now, in this case what it will mean is to accept the null hypothesis.

Thus, type II error will be to conclude the average transaction value is equal to $13.65, in reality the average transaction value is greater than $13.65.

constants mean the stopped points or the points on a graph that nothing is changing, hope this

step-by-step explanation:

so a parallelogram is just like a square base times height. multiply 40 and 39 to get 1560 inches.

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the kite can be split into two separate triangles and added together to get the total area. 30 x 4 /2 which equals 60. make sure to double it for the two triangles making 120.