Question

In: Statistics and Probability

1) A. In a Rasmussen Poll of 1020 adults in July 2010, 512 of those polled...

1) A. In a Rasmussen Poll of 1020 adults in July 2010, 512 of those polled said that schools should ban sugary snacks and soft drinks. Do a majority of adults support a ban on sugary snacks and soft drinks? Perform a hypothesis test using a significance level of 10%

B. Historically, the percentage of U.S. residents who support stricter gun control laws been 54%.A recent Gallup Poll of 1011 people showed 499 in favor of stricter gun 10% control laws. Assume the poll was given to a random sample of people. Test the claim that the proportion of those favoring stricter gun control has fallen. Perform a hypothesis test, using a significance level of 1%

Solutions

Expert Solution

A) Sample proportion saying schools should ban sugary snacks and soft drinks,   = 512/1020 = 0.502

We can do a single sample proportion Z-score test, where hypothesized population proportion, p0 = 0.5

Hypotheses:

H0: p <= 0.5

Ha: p > 0.5

=> Z-statistic, z =  ( - p0) / sqrt [p0 * (1-p0) / n] = 0.128

Right tailed Critical Z-score for significance level of 10%, Z-critical = Z0.1 = -1.28

Since z <= abs(Z-critical), we cannot reject the null hypothesis at the 10% significance level. Hence, majority of adults do not support the ban.

B) We perform one sample proportion Z-score test again:

H0: p >= 0.54

Ha: p < 0.54

Hypothesized population proportion, = 0.54

Sample proportion, p0 = 499 / 1011 = 0.494

=> Sample Z-statistic, z =  ( - p0) / sqrt [p0 * (1-p0) / n] = 2.938

Left tailed Critical Z-score for significance level of 1%, Z-critical = Z0.01 = 2.33

Since z > abs(Z-critical), we can reject the null hypothesis at the 1% significance level. Hence, the proportion of those favoring gun control has indeed fallen at the 1% significance level.


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