A survey of Internet users reported that 20% downloaded music onto their computers. The filing of lawsuits by the recording industry may be a reason why this percent has decreased from the estimate of 28% from a survey taken two years before. Assume that the sample sizes are both 1421. Using a significance test, evaluate whether or not there has been a change in the percent of Internet users who download music. Provide all details for the test. (Round your value for z to two decimal places. Round your P-value to four decimal places.)

Question

A survey of Internet users reported that 20% downloaded music onto their computers. The filing of lawsuits by the recording industry may be a reason why this percent has decreased from the estimate of 28% from a survey taken two years before. Assume that the sample sizes are both 1421. Using a significance test, evaluate whether or not there has been a change in the percent of Internet users who download music. Provide all details for the test. (Round your value for z to two decimal places. Round your P-value to four decimal places.)

z = ___________

Also report a 95% confidence interval for the difference in proportions. (Round your answers to four decimal places.) ( _____ , _____ )

Answer 1

my lungs will fill and then deflate they fill with fire exhaled desire i know its dirure my time today

Answer 2

To evaluate whether there has been a change in the percent of internet users who download music, we can use a significance test for comparing proportions.

Given:
- The new estimate is 20% (p1 = 0.20)
- The previous estimate is 28% (p2 = 0.28)
- The sample size for both surveys is 1421 (n1 = n2 = 1421)

Now, let's calculate the test statistic (z) using the following formula:

z = (p1 - p2) / sqrt((p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2))

Substituting the values into the formula:

z = (0.20 - 0.28) / sqrt((0.20 * (1 - 0.20) / 1421) + (0.28 * (1 - 0.28) / 1421))

Calculating the values within the square root:

z = (0.20 - 0.28) / sqrt((0.20 * 0.80 / 1421) + (0.28 * 0.72 / 1421))

z = -0.08 / sqrt(0.00011216064 + 0.0002507184)

z = -0.08 / sqrt(0.00036287904)

z = -0.08 / 0.01903668348

z ≈ -4.1964

The test statistic (z) is approximately -4.1964.

To determine the P-value associated with this test statistic, we need to compare it to the standard normal distribution. We can look up the corresponding P-value using a table or a statistical calculator. The P-value represents the probability of obtaining a test statistic as extreme as the observed result (or even more extreme) under the null hypothesis.

The P-value for z ≈ -4.1964 is around 0.0000 (very close to zero).

Since the P-value is less than our significance level (usually 0.05), we would reject the null hypothesis. This means there is evidence to suggest a change in the percent of Internet users who download music.

Next, let's calculate the confidence interval for the difference in proportions. We can use the following formula:

p1 - p2 ± z * sqrt((p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2))

Substituting the values into the formula:

Interval = 0.20 - 0.28 ± 1.96 * sqrt((0.20 * 0.80 / 1421) + (0.28 * 0.72 / 1421))

Calculating the values within the square root:

Interval = 0.20 - 0.28 ± 1.96 * sqrt(0.00011216064 + 0.0002507184)

Interval = 0.20 - 0.28 ± 1.96 * sqrt(0.00036287904)

Interval = 0.20 - 0.28 ± 1.96 * 0.01903668348

Interval = -0.08 ± 1.96 * 0.01903668348

Interval = -0.08 ± 0.03736449726

The 95% confidence interval for the difference in proportions is approximately (-0.1174, -0.0426).

Therefore, we can conclude that, based on the significance test and the confidence interval, there is evidence of a decrease in the percent of Internet users who download music, and the estimated difference in proportions falls within the interval (-0.1174, -0.0426).