Question

Using the current system, the mean wait time for people who call for technical support is 5 minutes. A new system was designed to reduce the mean wait time. A random sample of 400 calls produced a sample average of 4.650. The standard deviation of wait times is known to be 4. We want to know whether the new software changed the mean.

1. What is the null hypothesis and what is the alternative hypothesis?

2. What is the value of the test statistic?

3. What is the p-value of the test? (Compute to 4 digits).

4. Do we reject H0 at the .05 level of significance?

5. Do we reject H0 at the .01 level of significance?

Answer 1

Solution :

Given that,

1)

This is a two tailed test.

The null and alternative hypothesis is,

Ho: $\mu$ $=$ 5

Ha: $\mu$ $\neq$ 5

2)

The test statistics,

Z =( $\bar x$ - $\mu$ )/ ($\sigma$/$\sqrt{}$n)

= ( 4.650 - 5 ) / ( 4 /$\sqrt{}$400 )

= -1.75

3)

P- Value = 2*P(Z < z )

= 2 * P(Z < -1.75 )

= 2 * 0.0401

= 0.0802

4)

The p-value is p = 0.0802, and since p = 0.0802 $\geq$ 0.05, it is concluded that do not reject the null hypothesis at the 0.05 level of significance.

5)

The p-value is p = 0.0802, and since p = 0.0802 $\geq$ 0.01, it is concluded that do not reject the null hypothesis at the 0.01 level of significance.