Question

A researcher from a construction firm claims that the average cost of total renovation of a bungalow house is less than $85 (in ten thousand dollars). He selects a random sample of 36 pairs of shoes from a catalog and finds the following costs (in ten thousand dollars) 62 70 75 55 80 55 51 40 80 70 52 96 120 90 75 85 80 60 110 65 82 85 85 45 75 60 90 92 60 95 110 85 45 90 70 71. Is there enough evidence to support the researcher's claim at a 0.05? Assume σ=18.2.

Answer 1

Given that, sample size ( n ) = 36

population standard deviation $(\sigma)$ = 18.2

significance level $(\alpha)$ = 0.05

Let X be costs ( in ten thousand dollars)

Mean value of X is,

$\bar x = \frac { \sum X }{ n }= \frac { 2711}{36} = 75.3056$

The null and the alternative hypotheses are,

$H_0: \mu = \$ 85$

$H_a: \mu < \$ 85$

Test statistic is,

$Z = \frac { \bar x -\mu}{\frac { \sigma}{\sqrt {n}}} = \frac { 75.3056- 85}{\frac { 18.2}{\sqrt { 36}}} = -3.20$

Test statistic is, z = -3.20

p-value = P(Z < -3.20) = 0.0007

p-value = 0.0007

Here, p-value = 0.0007 < $\alpha = 0.05$

So, we reject the null hypothesis ( H₀).

Conclusion: there is sufficient evidence to support the researcher's claim that the average cost of total renovation of a bungalow house is less than $85 (in ten thousand dollars).