Question

A random sample of data was obtained 61 12 6 40 27 38 93 5 13 40

a) How many data items? n = ?

b) What is the sample mean?

c) What is the sample variance?

d) What is the sample standard deviation?

e) Based on the number of sample items will you be using a z or a t for the mean testing?

f) Form a 95% confidence interval for the mean.

g) Form a 97% confidence interval for the variance.

h) Use Hypothesis Testing (P-Value Method) for the claim that the mean is less than 30 at a 96% confidence level.

i) What is your conclusion for the Hypothesis Test, state it very clearly?

Answer 1

a) How many data items?

n = 10

b) What is the sample mean?

33.5

c) What is the sample variance?

766.056

d) What is the sample standard deviation?

27.678

e) Based on the number of sample items will you be using a z or a t for the mean testing?

A t

f) Form a 95% confidence interval for the mean.

33.500	mean Data
27.678	std. dev.
8.752	std. error
10	n
9	df
13.701	confidence interval 95.% lower
53.299	confidence interval 95.% upper
19.799	margin of error

g) Form a 97% confidence interval for the variance.

766.055556	observed variance of 1
10	n
9	df
336.112	confidence interval 97.% lower
2,952.858	confidence interval 97.% upper

h) Use Hypothesis Testing (P-Value Method) for the claim that the mean is less than 30 at a 96% confidence level.

The hypothesis being tested is:

H₀: µ = 30

H_a: µ < 30

30.000	hypothesized value
33.500	mean Data
27.678	std. dev.
8.752	std. error
10	n
9	df
0.400	t
.6507	p-value (one-tailed, lower)

Since the p-value (0.6507) is greater than the significance level (0.04), we cannot reject the null hypothesis.

i) What is your conclusion for the Hypothesis Test, state it very clearly?

Therefore, we cannot conclude that the mean is less than 30 at a 96% confidence level.