Question

The Iowa Department of Transportation repaired hundreds of bridges in 1993. To check the average cost to repair a bridge, a random sample of n = 55 bridges were chosen. The mean and standard deviation for the sample are $25,788 and $1,540, respectively. Records from previous years indicate an average bridge repair cost was $25,003. Use the sample data to test that 1993 mean μ is greater than $25,003. Use α = 0.05. What type of test would be used?

Answer 1

We conclude that the average cost to repair a bridge is greater than $25,003.

Explanation:

We are given that to check the average cost to repair a bridge, a random sample of n = 55 bridges were chosen. The mean and standard deviation for the sample are $25,788 and $1,540, respectively.

Records from previous years indicate an average bridge repair cost was $25,003.

Let
`\mu` = average cost to repair a bridge.

So, Null Hypothesis,
`H_0` :
`\mu \leq` $25,003 {means that the average cost to repair a bridge is smaller than or equal to $25,003}

Alternate Hypothesis,
`H_A` :
`\mu` > $25,003 {means that the average cost to repair a bridge is greater than $25,003}

The test statistics that would be used here One-sample t-test statistics as we don't know about population standard deviation;

T.S. =
`(\bar X-\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`\bar X` = sample mean cost to repair a bridge = $25,788

s = sample standard deviation = $1,540

n = sample of bridges = 55

So, the test statistics =
`(25,788-25,003)/((1,540)/(√(55) ) )` ~
`t_5_4`

= 3.78

The value of t test statistic is 3.78.

Now, at 5% significance level the t table gives critical value of 1.674 at 54 degree of freedom for right-tailed test.

Since our test statistic is more than the critical value of t as 3.78 > 1.674, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the average cost to repair a bridge is greater than $25,003.