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Natalka [10]
3 years ago
8

To get an estimate of consumer spending over a holiday season in 2009, 436 randomly sampled American adults were surveyed. Their

spending for the six-day period after Thanksgiving, spanning the Black Friday weekend and Cyber Monday, averaged $84.71. A 95% confidence interval based on this sample is ($80.31, $89.11). Which of the following statements are true? Select all that apply. A 90% confidence interval would be narrower than the 95% confidence interval if we don't need to be as sure about our estimate. In order to decrease the margin of error of a 95% confidence interval to a third of what is is now, we would need to use a sample 3 times larger. This confidence interval is not valid since the distribution of spending in the sample data is right skewed. The margin of error is $4.4. We are 95% confident that the average spending of these 435 American adults over this holiday season is between $80.31 and $89.11. This confidence interval is valid since the sampling distribution of sample mean would be approximately normal with sample size of 436. 95% of random samples have a sample mean between $80.31 and $89.11. We are 95% confident that the average spending of all American adults over this holiday season is between $80.31 and $89.11.
Mathematics
1 answer:
Tatiana [17]3 years ago
6 0

Answer:

<em>-A 90% confidence interval would be narrower than the 95% confidence interval if we don't need to be as sure about our estimate. </em>

<em>-This confidence interval is not valid since the distribution of spending in the sample data is right skewed.</em>

<em>-The margin of error is $4.4.</em>

<em>-This confidence interval is valid since the sampling distribution of sample mean would be approximately normal with sample size of 436.</em>

<em>-We are 95% confident that the average spending of all American adults over this holiday season is between $80.31 and $89.11.</em>

<em />

Step-by-step explanation:

<em>A 90% confidence interval would be narrower than the 95% confidence interval if we don't need to be as sure about our estimate. </em>

TRUE. The 90% confidence is less strict in its probability of having the mean within the interval, so it is narrower than the 95% CI. It relies more in the information given by the sample.

<em />

<em>In order to decrease the margin of error of a 95% confidence interval to a third of what is is now, we would need to use a sample 3 times larger. </em>

FALSE. The margin of error is z*σ/(n^0.5). So to reduce it by two thirds, the sample size n needs to be 3^2=9 times larger.

<em>This confidence interval is not valid since the distribution of spending in the sample data is right skewed.</em>

FALSE. There is no information about the skewness in the sample.

<em>The margin of error is $4.4.</em>

TRUE. The margin of error is (89.11-80.31)/2=$4.4.

<em>We are 95% confident that the average spending of these 435 American adults over this holiday season is between $80.31 and $89.11.</em>

FALSE. The CI is related to the populations mean. We are 95% confident that the average spending of the population is between $80.31 and $89.11.

<em>This confidence interval is valid since the sampling distribution of sample mean would be approximately normal with sample size of 436.</em>

TRUE. This happens accordingly to the Central Limit Theorem.

<em>95% of random samples have a sample mean between $80.31 and $89.11.</em>

FALSE. The confidence interval refers to the population mean.

<em>We are 95% confident that the average spending of all American adults over this holiday season is between $80.31 and $89.11.</em>

TRUE. This is the conclusion that is looked for when constructing a confidence interval.

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This question is incomplete, the complete question is;

In a survey, 55% of the voters support a particular referendum. If 40 voters are chosen at random,

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b) Variance is 9.9

Step-by-step explanation:

Given that;

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