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.

Null hypothesis is

Alternative hypothesis is

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

Step-by-step explanation:

Given that:

Mean = 960 hours

Sample size n = 36

Mean population 937

Standard deviation = 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

Alternative hypothesis is

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 = 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

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step-by-step explanation:

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