Question

The CEO of a large electric utility claims that 80 percent of his 1,000,000 customers are very satisfied with the service they receive. To test this claim, the local newspaper surveyed 100 customers, using simple random sampling Among the sampled customers, 74 percent say they are very satisfied.
a)[1 pt] Is this a right-tailed, left-tailed, or 2 -tailed test?
b) [2 pts] Find the value of the test statistic =
b) [2 pts]Find the p-value [round to 3 decimals]
c) I[1 pt]If the level of significance is 5%, what is your decision? Explain
d){1 pt] Do the statistics support the CEO's claim that 80% are satisfied? Explain.

Answer 1

Here we have given that,

n=number of local newspaper customers=100

=Sample proporiton of local newspaper customers very satisfied with the service they receive=74%=0.74

(a)

Claim: To check whether the population proportion of the customers are very satisfied with the service they receive is equal to 80% i.e. 0.80.

The null and alternative hypothesis are as follows,

Versus

where p is the population proportion of the customers are very satisfied with the service they receive

This is two tailed test.

(b)

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

Z-statistics=

=

= -1.50

The test statistics is -1.50.
Now we find the P-value

p-value=2*( P(Z < z-statistics)) as this two tailed test

             =2* P( Z < -1.50)

             =2* (0.06618) Using standard normal z table see the value corresponding to the z=-1.50

             = 0.1324

The p-value is 0.1324

(c)

Decision:

= level of significance= 0.05

Here, p-value (0.1324) greater than (>) 0.05

We fall to reject the Ho (Null Hypothesis)

(d)

Conclusion:

Here, There is sufficient evidence to support the claim that the population proportion of the customers are very satisfied with the service they receive is equal to 80% i.e. 0.80.

we can conclude that the population proportion of the customers are very satisfied with the service they receive is equal to 80% i.e. 0.80.