According to a Gallup poll conducted in March 2001 63 of Ame

According to a Gallup poll conducted in March 2001, 63% of Americans were either fairly
concerned or greatly concerned about global warming. The same poll question was asked of
1021 randomly selected adults a decade later, and a sample proportion of the people fairly
concerned or greatly concerned about global warming was 51%.

((a) Calculate a 95% confidence interval for the true proportion of Americans who were either
fairly concerned or greatly concerned about global warming in 2011.
(b) Do we have enough evidence to say that Americans are worrying less about environmental
issues than in 2001? State the hypotheses that you will test.
(c) Find the test statistic and the P-value.
(d) Sketch a standard Normal curve with the area corresponding to the P-value shaded.
(e) Is the result statistically significant at the = 0.05 level? Why or why not?
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(f) Do you think there is convincing evidence that Americans are worrying less about
environmental issues? Give a reason for your answer.

A test of the null hypothesis H0: p = .50 gives test statistic z = 1.8.
(a) What is the P-value if the alternative is Ha: p > .50?
(b) What is the P-value if the alternative is Ha: p < .50?
(c) What is the P-value if the alternative is Ha: p .50?
Finally, in your own words, what is a p-value?

Solution

We have given that 63% of Americans were either fairly concerned or greatly concerned about global warming.

The same poll question was asked of 1021 randomly selected adults a decade later, and a sample proportion of the people fairly concerned or greatly concerned about global warming was 51%.

Here proportion of Americans were either fairly concerned or greatly concerned about global warming

P = 63% = 0.63 which is a population proportion.

The people fairly concerned or greatly concerned about global warming

Sample proportion (p) = 51% = 0.51

(a) Calculate a 95% confidence interval for the true proportion of Americans who were either fairly concerned or greatly concerned about global warming in 2011.

A 95% confidence interval for proprtion of Americans who were either fairly concerned or greatly concerned about global warming in 2011 that is for P(population proportion) is,

p - E < P < p + E

Where p is sample proportion

E is margin of error.

E = Z_c * sqrt(p*q / n)

p = 0.51 q = 1 - p = 1 - 0.51 = 0.49

We will calculcte Z_c by using EXCEL.

c = confidence level = 95% = 95/100 = 0.95

= 1 - c = 1 - 0.95 = 0.05

/2 = 0.05 / 2 = 0.025

1 - /2 = 1 - 0.025 = 0.975

We will use EXCEL command as,

= NORMSINV(probability = 0.975)

Which gives us Z_c = 1.96

sample size (n) = 1021

E = 1.96 * [ (0.51 * 0.49) / 1021 ] = 1.96 * 0.01564 = 0.03065

Lower limit = p - E = 0.51 - 0.03065 = 0.47935

Upper limit = p + E = 0.51 + 0.03065 = 0.54065

The confidence interval is (0.47935 , 0.54065).

(b) Do we have enough evidence to say that Americans are worrying less about environmental issues than in 2001? State the hypotheses that you will test.

Here we have to test the hypothesis that,

H₀ : P = 0.63 Vs H₁ : P < 0.63

The test statistic for testing this hypothesis is,

Z = ( p - P ) / sqrt [ (p*q) / n ]

= ( 0.51 - 0.63) / sqrt [ (0.51 * 0.49) / 1021 ]

= - 0.12 / 0.01564

Z = - 7.673

P-value : The test is left sided so p-value of the left tail test is area to the left of test statistic (for Z = -7.673)

The p-value is very very small that is 8.40097 * 10^-15 which is nearly equal to 0.

Given that = 0.05

Decision : p-value < 0.05

Reject H₀ at 5% level of significance.

Conclusion : There is enough evidence to say that Americans are worrying less about environmental
issues than in 2001.

A test of the null hypothesis H0: p = .50 gives test statistic z = 1.8.

(a) What is the P-value if the alternative is Ha: p > .50?

Here we have given that test statistic is, z = 1.8

p-value is the area to the right of the test statistic (z = 1.8).

p-value = 1 - 0.96407 = 0.03593

(b) What is the P-value if the alternative is Ha: p < .50?

p-value is the area to the left of the test statistic (z = 1.8)

p-value = 0.03593

Because the normal curve is symmetric so that both the areas (to the right and to the left) are same.

(c) What is the P-value if the alternative is Ha: p .50?

p-value is the area to the left and to the right.

p-value = area to the left + area to the right = 0.03593 + 0.03593 = 0.071861