Question

In tests of a computer component, it is found that the mean time between failures is 937 hours. A modification is made which is supposed to increase reliability by increasing the time between failures. Tests on a sample of 36 modified components produce a mean time between failures of 960 hours, with a standard deviation of 52 hours. Using a significance level of .01, test the claim that, for modified components, the mean time between failures is greater than 937 hours. Find the appropriate p-value.

Answer 1

Null hypothesis is
`\mathbf {H_o: \mu > 937}`

Alternative hypothesis is
`\mathbf {H_a: \mu < 937}`

Test Statistics z = 2.65

CONCLUSION:

Since test statistics is greater than critical value; we reject the null hypothesis. Thus, there is sufficient evidence to support the claim that the modified components have a longer mean time between failures.

P- value = 0.004025

Explanation:

Given that:

Mean
`\overline x` = 960 hours

Sample size n = 36

Mean population
`\mu =` 937

Standard deviation
`\sigma` = 52

Given that the mean time between failures is 937 hours. The objective is to determine if the mean time between failures is greater than 937 hours

Null hypothesis is
`\mathbf {H_o: \mu > 937}`

Alternative hypothesis is
`\mathbf {H_a: \mu < 937}`

Degree of freedom = n-1

Degree of freedom = 36-1

Degree of freedom = 35

The level of significance ∝ = 0.01

SInce the degree of freedom is 35 and the level of significance ∝ = 0.01;

from t-table t(0.99,35), the critical value = 2.438

The test statistics is :

`Z = (\overline x - \mu )/((\sigma)/(√(n)))`

`Z = (960-937 )/((52)/(√(36)))`

`Z = (23)/(8.66)`

Z = 2.65

The decision rule is to reject null hypothesis if test statistics is greater than critical value.

CONCLUSION:

Since test statistics is greater than critical value; we reject the null hypothesis. Thus, there is sufficient evidence to support the claim that the modified components have a longer mean time between failures.

The P-value can be calculated as follows:

find P(z < - 2.65) from normal distribution tables

= 1 - P (z ≤ 2.65)

= 1 - 0.995975 (using the Excel Function: =NORMDIST(z))

= 0.004025