Question

a)

Suppose we have a left-tailed hypothesis test about µ conducted at α = 0.10. If the sample size is n = 7, what is the correct rejection region?

Group of answer choices

t > 1.440

z > 1.28

z < -1.28

t < -1.440

b)

Danny was recently hired by Mr. Peanut to demonstrate (at α = 0.05) that less than 50% of Mr. Peanuts’ Mixed Nuts are peanuts. Danny randomly sampled 120 mixed nuts and found that 42 were peanuts. Then Danny contracted coronavirus and could not compute the test statistic. Will you please figure out the test statistic?

Group of answer choices

42

3.29

0.05

-3.29

c)

We intend to estimate the average break time of Dunder Mifflin employees. From a previous study, we believe that the average time is 42 minutes with a standard deviation of 9 minutes. We want our 95 percent confidence interval to have a margin of error of no more than plus or minus 3 minutes. What is the smallest sample size that we should consider?

Group of answer choices

4

104

6

35

e)

UCF administrators want to determine the proportion of students who have seen The Matrix. If the proportion differs from 60%, then the university will be forced to permanently close. Suppose a hypothesis test is conducted and the test statistic is -1.23. What is the p-value for a two-tailed test of hypothesis?

Group of answer choices

0.2186

0.7814

0.3907

0.1093

Answer 1

A)

As the sample size is small we will use t distribution table to estimate the rejection region

N = 7

Degrees of freedom is = n-1 = 6

For 6 dof and 0.1 alpha, critical value t from t table is -1.44

Rejection region is

t < -1.44

B)

Null hypothesis Ho : P = 0.5

Alternate hypothesis Ha : P < 0.5

N = 120

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 60

N*(1-p) = 60

Both the conditions are met so we can use standard normal z table to estimate the P-Value

Test statistics z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

Observed P = 42/120

Claimed P = 0.5

N = 120

Z = -3.29