Question

True or False? In observing 9 workers assembling similar devices, the manager noticed that their average time taken was 16.2 minutes with a standard deviation of 3.6. At a 10% level of significance we fail to reject the null hypothesis that the true mean equals 15 minutes, versus the alternate hypothesis that the true mean exceeds 15.

True or False? A politician claims to have support from more than 50% of the population. In a opinion poll, 648 people out of 1200 supported the politician. At a 1% level of significance we conclude that the true proportion p is  more than 50%.

Answer 1

(A)

Here we have given that,

Xi: Time minutes

n=number of workers =9

= sample mean time = 16.2 minutes

S= sample standard deviation= 3.6 minutes

Here, the population standard deviation is not known we are using the one-sample t-test to test the hypothesis.

Claim: To check whether the population mean time exceeds 15.

The null and alternative hypothesis is as follows

Versus

where = Population mean

This is the two-tailed test.

Now, we can find the test statistic

t-statistics =

                  =

                  =

                   =1.00

The Test statistic is 1.00

Now we can find the P-value

Degrees of freedom = n-1 = 9-1 =8

This is the right one-tailed test

P-value =0.1733   Using Excel software= TDIST ( |t-statistics | = 1.00, Degrees of freedom =8, tail=1)

we get the P-value is 0.1733

Decision:

= level of significance=0.10

Here, P-value (0.1733) greater than (>) 0.10 ()

Conclusion:

we fail to reject Ho (Null Hypothesis)

There is no sufficient evidence to support the claim the population mean time exceeds 15.

i.e. the population mean time is equal to 15.

(B)

Here we have given that,

n=number of people=1200

x: number of people supported the politician =648

Now, we estimate the sample proportion as

=sample proportion =

Claim: To check whether the true population proportion of people supported the politician is more than 50% i.e. 0.50.

The null and alternative hypothesis are as follows,

Versus

where p is the true population proportion of people supported the politician=0.50

This is the right one-tailed test.

Now, we can find the test statistic is as follows,

Z-statistics=

=

=

   =2.77

The test statistics is 2.77.

Now we find the P-value,

p-value=P(Z > z-statistics) as this is right one-tailed test

              =1- P( Z < 2.77)

=1 - 0.99720 Using standard normal z table see the value corresponding to the z=2.77

              =0.0028

The p-value is 0.0028

Decision:

= level of significance= 0.01

Here p-value (0.0028) less than (<) 0.01

Conclusion:

We reject the Ho (Null Hypothesis)

There is sufficient evidence to support the claim the true population proportion of people supported the politician is more than 50% i.e. 0.50.