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.

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

-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.

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

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

Step-by-step explanation:

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

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.

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.

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.

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

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

The margin of error is $4.4.

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

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.

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

TRUE. This happens accordingly to the Central Limit Theorem.

95% of random samples have a sample mean between $80.31 and $89.11.

FALSE. The confidence interval refers to the population mean.

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

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

no, the equation would be invalid since they are both not equal.

step-by-step explanation:

here you got

4y-6+4y 2-9

first, you would simplify the equation on the form.

4y-6+4y 2-9

then, move all the terms that contains a y to the left and all the other terms to the right.

+4y-4y+-9-6

finally, you will simplify the left and the right side of the equation.

y=-3