Question

A recent survey indicated that the average amount spent for breakfast by business managers was $7.58 with a population standard deviation of $0.42. It was felt that breakfasts on the West Coast were higher than $7.58. A sample of 81 business managers on the West Coast had an average breakfast cost of $7.65. At the .05 significance level, is the price of breakfasts on the West Coast significantly higher?

a) State the hypotheses and identify the claim

b) Find the critical values

c) Compute the test values

d) Make the decision

e) Summarize the results

Answer 1

Concepts and reason

The concept of one sample z-test is used to solve this problem.

The parametric tests used to test the population mean $\left( \mu \right)$ of one variable are z test and t test.

The z test is used when the standard deviation of the population $\left( \sigma \right)$ is known.

The t test is used when the standard deviation of the population $\left( \sigma \right)$ is unknown.

Fundamentals

The test statistics is calculated as,

$z = \frac{{\bar x - \mu }}{{\frac{\sigma }{{\sqrt n }}}}$

Where $\bar x$ is the mean of sample, $\mu$ is the mean of population, $\sigma$ is the standard deviation of the population, and $n$ is the size of sample.

(a)

The claim of the study is to test that the price of breakfasts on the West Coast is significantly higher than the population mean $7.58. So, the hypothesis will be framed as,

Null hypothesis:

${H_0}:\mu = 7.58$

Alternative hypothesis:

${H_1}:\mu > 7.58$

(b)

The critical value for the test would be depend on the significance level. The level of significance used is 0.05.

The critical value of Z is obtained by using the standard normal table at 5 % level of significance, as 1.645.

(c)

The provided information is,

For men,

$\begin{array}{c}\\{\rm{ sample size}}\left( n \right) = 81\\\\{\rm{ sample mean}}\left( {\bar x} \right) = 7.65\\\\{\rm{population standard deviation}}\left( \sigma \right) = 0.42\\\end{array}$

The test statistics is calculated as,

$\begin{array}{c}\\z = \frac{{\bar x - \mu }}{{\frac{\sigma }{{\sqrt n }}}}\\\\ = \frac{{7.65 - 7.58}}{{\frac{{0.42}}{{\sqrt {81} }}}}\\\\ = 1.5\\\end{array}$

(d)

The value of test statistic is 1.5 and the critical value is 1.645.

Since the test statistic value 1.5 is less than the critical value 1.645, so the researcher is failed to reject the null hypothesis.

(e)

Since the decision is failed to reject the null hypothesis as the test statistic 1.5 is less than the critical value 1.645, therefore researcher concludes that there is no significant increase in the price of breakfasts on the West Coast as compared to population mean $7.58

Ans: Part a

The null and the alternative hypothesis is,

$\begin{array}{l}\\{H_0}:\mu = 7.58\\\\{H_1}:\mu > 7.58\\\end{array}$

Part b

The critical value for the test is 1.645.

Part c

The value of the z test statistic is 1.5.

Part d

The decision is that the null hypothesis is accepted.

Part e

The price of breakfast does not increase significantly as compared to population mean $7.58.