Question

According to the Census Bureau, 3.29 people reside in the typical American household. A sample of 16 households in Arizona retirement communities showed the mean number of residents per household was 2.76 residents. The standard deviation of this sample was 1.29 residents. At the 0.05 significance level, is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.29 persons?

(a) State the null hypothesis and the alternate hypothesis. (Round your answer to 2 decimal places.)
H0: μ ≥
H1: μ <
(b) State the decision rule for .05 significance level. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)
Reject H0 if t <
(c) Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)
Value of the test statistic
(d) Is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.29 persons
(Click to select)Fail to rejectReject H0. Mean number of residents (Click to select)is necessarilyis not necessarily less than 3.29 persons.

Answer 1

Kindly check explanation

Explanation:

Given the following :

Alpha = 0.05

Sample size (n) = 16

Population mean(σ) = 3.29

Sample mean (m) = 2.76

Sample standard deviation (sd) = 1.29

(1)

Null hypothesis : μ ≥ 3.29

Alternative hypothesis : μ < 3.29

(2)

From the t distribution table :

The Null hypothesis should be rejected if

t < - t0

To obtain t0:

Degree of freedom (df) = n - 1 = 16 - 1 = 15 at alpha = 0.05

t0 = 1.753 ( left tailed) = - 1.753

(3)

Value of test statistic:

t = (m - σ) / (sd / √n)

(2.76 - 3.29) / (1.29 / √16)

-0.53 / 0.3225

t = - 1.643

Comparing t < - t0

But : -1.643 > - 1.753

Hence,

t > - t0

Hence, we fail to reject the null hypothesis