Question

In: Statistics and Probability

Bodacious Burger, Inc., claims that the average weight of its “Pi Over Twelve” burger is 0.261799...

Bodacious Burger, Inc., claims that the average weight of its “Pi Over Twelve” burger is 0.261799 lb with a standard deviation of 0.000004 lb. Jorge, however, has weighed 42 of these burgers and found a mean weight of 0.261800 lb, and intends to test the company’s claim using this data. Perform the test for him at the 0.05 significance level and answer the questions below.


Jorge’s null hypothesis is:

A. μ≠0.261799
B. μ=0.261799
C. μ≤0.261799
D. μ≥0.261799

To two decimal places, p=


Conclusion:

A. Reject the null hypothesis
B. Do not reject the null hypothesis

To test the hypotheses H0:μ≥210.21, Ha:μ<210.21 for a normal population at the α=0.05 significance level, a sample of size 23 was drawn and found to have mean x¯=198.73 and standard deviation s=19.10.

Which of the following is closest to the p-value that was found? (Assume that the p-value was found correctly.)

A. 0.00
B. -0.01
C. 0.02
D. 0.01
E. -0.02

Which of the following is the correct conclusion from the test?

A. The test should not be performed
B. Reject the null hypothesis
C. Do not reject the null hypothesis

Kreem, Inc. claims that its ice cream contains an average of 225.00 calories per half-cup serving. To test that claim, Jeanette has measured the calories in 43 servings, for which she found a mean of 229.11 and a standard deviation of 17.60. The correct conclusion from the hypothesis test she performs is:

A. Do not reject the null hypothesis at the 0.01 significance level
B. The test should not be performed
C. Reject the null hypothesis at the 0.01 significance level
D. Reject the null at the 0.05 significance level

Jodi has conducted a hypothesis test for H0:μ≤66 vs. Ha:μ>66, found a p-value of 0.12, and decided not to reject the null hypothesis. What conclusion is associated with this decision?

A. There is not good evidence to show that μ>66.
B. There is good evidence to show that μ>66.
C. There is good evidence to show that μ≤66.
D. There is not good evidence to show that μ≤66.

When the hypotheses Ha:μ≠200.61 vs H0:μ=200.61 are tested at the 0.1 level of significance, given that a sample of size 42 had a mean of 191.47 and a standard deviation of 35.4, the correct conclusion is:

A. p=0.1, reject the null hypothesis
B. p=0.1, do not reject the null hypothesis
C. p=0.09, reject the null hypothesis
D. p=0.09, do not reject the null hypothesis

When the hypotheses Ha:μ≠205.99 vs H0:μ=205.99 are tested at the 0.1 level of significance, given that a mean of 196.59 was found for a sample of size 40 taken from a population with σ=43.17, the correct conclusion is:

A. p=0.17, do not reject the null hypothesis
B. p=0.18, reject the null hypothesis
C. p=0.17, reject the null hypothesis
D. The test should not be performed
E. p=0.18, do not reject the null hypothesis

Solutions

Expert Solution

1)


Null Hypothesis, H0: μ = 0.261799

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (0.2618 - 0.261799)/(0.000004/sqrt(42))
z = 1.62

P-value Approach
P-value = 0.11

B. Do not reject the null hypothesis

2)

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (198.73 - 210.21)/(19.1/sqrt(23))
t = -2.883

P-value Approach
P-value = 0.00

As P-value < 0.05, reject the null hypothesis.


3)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 225
Alternative Hypothesis, Ha: μ ≠ 225

Rejection Region
This is two tailed test, for α = 0.01 and df = 42
Critical value of t are -2.698 and 2.698.
Hence reject H0 if t < -2.698 or t > 2.698

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (229.11 - 225)/(17.6/sqrt(43))
t = 1.531

P-value Approach
P-value = 0.13
As P-value >= 0.01, fail to reject null hypothesis.

A. Do not reject the null hypothesis at the 0.01 significance level

4)

A. There is not good evidence to show that μ>66.

5)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 200.61
Alternative Hypothesis, Ha: μ ≠ 200.61

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (191.47 - 200.61)/(35.4/sqrt(42))
t = -1.673

P-value Approach
P-value = 0.1
As P-value >= 0.1, fail to reject null hypothesis.


B. p=0.1, do not reject the null hypothesis

6)

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (196.59 - 205.99)/(43.17/sqrt(40))
z = -1.38

P-value Approach
P-value = 0.17
As P-value >= 0.1, fail to reject null hypothesis.


A. p=0.17, do not reject the null hypothesis


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