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 to determine if the proportion differs from 80%. Among the sampled customers, 73 say they are very satisfied. Based on these findings, can we reject the CEO's hypothesis that 80% of the customers are very satisfied? Use a 0.05 level of significance.

a) State the null hypothesis and the alternative hypothesis.

            b) State the test that should be used to test this hypothesis?

            c) Compute the appropriate test statistic for this test.

            d) Determine the critical value(s) for the significance level.

            e) Determine the p-value for this test.

            f) Based on the p-value or critical value(s) you found above, what is your conclusion. Explain.

            g) Find a 95% confidence interval and explain how this supports your conclusion.

Answer 1

ANSWER:

Given,

x = 73

n = 100

$\small \widehat{p}$ = x / n

= 73/100

= 0.73

a)

Null hypothesis Ho : p = 0.80

Alternative hypothesis Ha : p $\neq$ 0.80

b)

The test that should be used to test this hypothesis is Z- test

c)

Test statistic(Z):

$\small Z=\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$\small =\frac{0.73-0.80}{\sqrt{\frac{0.80*0.20}{100}}}$

$\small =\frac{-0.07}{0.04}$

Z = -1.75

d)

significance level($\small \alpha$) =0.05

Z_critical = Z_0.05 = 1.96 = Z

e)

At Z = -1.75

P - value = 0.0401

f)

P -value < $\small \alpha$

0.0401 < 0.05

Reject H₀.,so we can conclude that the 80% are not satisfied.

g)

95% confidence interval (C.I):

$\small C.I =\widehat{p}\pm Z\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

$\small =0.73\pm 1.96\sqrt{\frac{0.73*0.27}{100}}$

= 0.73 $\small \pm$ 0.0870

C.I = (0.643 , 0.817)

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