Question

According to a survey, 56% of young Americans aged 18 to 29 say the primary way they watch television is through streaming services on the internet. Suppose a random sample of 350 Americans from this group is selected. Would it be surprising to find that more than 70% of the sample watched television primary through streaming services? (Include z-score)

Answer 1

To determine if it would be surprising to find that more than 70% of the sample watched television primarily through streaming services, we can calculate the z-score and compare it to the standard normal distribution.

Step-by-step explanation:

To determine if it would be surprising to find that more than 70% of the sample watched television primarily through streaming services, we can calculate the z-score and compare it to the standard normal distribution.

1. First, we need to calculate the standard deviation of the sample proportion, which can be done using the formula: standard deviation = sqrt((p-hat*(1-p-hat))/n), where p-hat is the sample proportion, n is the sample size, and sqrt is the square root function.
2. Next, we calculate the z-score using the formula: z = (observed proportion - hypothesized proportion)/standard deviation. In this case, the hypothesized proportion is 70%.
3. Finally, we can look up the z-score in a standard normal distribution table to find the corresponding p-value. If the p-value is less than the significance level (e.g., 0.05), we would consider the result to be statistically significant and therefore surprising.

Answer 2

The z-score is a statistical measure that describes how far a sample estimate is from the population parameter in standard deviation units. It is calculated by taking the difference between the sample estimate and the population parameter, and dividing it by the standard deviation of the sampling distribution of the estimate. The resulting z-score can be used to calculate the probability of observing a sample estimate as extreme or more extreme than the one obtained, assuming the null hypothesis is true.

In this case, the sample estimate is the proportion of individuals who watch television primarily through streaming services, which is 70%, and the population parameter is the proportion of all individuals who watch television primarily through streaming services, which is 56%. The sample size is 350. The null hypothesis is that there is no difference between the sample and population proportions, and the alternative hypothesis is that the sample proportion is greater than the population proportion.

To calculate the z-score, we need to determine the standard error of the sampling distribution of the proportion. The standard error is a measure of the variability of sample estimates due to random sampling variation. It is calculated by taking the square root of the population proportion times one minus the population proportion, divided by the sample size. In this case, the standard error is:

standard error = sqrt[(0.56 * (1 - 0.56)) / 350] = 0.028

The z-score can now be calculated as:

z-score = (0.7 - 0.56) / 0.028 = 5.19

This z-score is quite large, indicating that the sample estimate is more than 5 standard errors away from the population parameter. This suggests strong evidence against the null hypothesis, as the probability of observing a sample estimate as extreme or more extreme than the one obtained is very low, assuming the null hypothesis is true.

In other words, the result of observing a sample proportion of 70% or above is highly unlikely to have occurred by chance alone, assuming that the true population proportion is 56%. This result provides evidence in support of the alternative hypothesis, which states that the sample proportion is greater than the population proportion.