The 2009 holiday retail season which kicked off on November 27, 2009, had been m

Question

The 2009 holiday retail season which kicked off on November 27, 2009, had been marked by somewhat lower self-reported consumer spending than was seen during the comparable period in 2008. To get an estimate of consumer spending 436 randomly sampled American adults were surveyed. Daily consumer spending on average based on this sample for the six-day period after Thanksgiving, spanning the black friday weekend and cyber monday has a 95% confidence interval, which is ($70, $78). Determine whether the following statements are true or false based on the given info. Please justify your response.

1. This confidence interval uses the result of the central limit theorem when calculating the standard error T/F Justify

2. We are 95% confident that the average spending of all American adults is between $70 and $78 with being point estimate of $75 T/F Justify

3. The margin of error is 4 and the standard error is around 2 T/F Justify

4. The margin of error in a 90% confidence interval is 2 T/F Justify

5. The standard deviation of the sample can be calculated with the given info T/F Justify

Please show all steps and why it is either true or false! Thank you!!

Explanation / Answer

Answer to the question)

The sample size = 436

It is a large sample

confidence level = 95%

confidence limit = (70,78)

Point estimate = (Lower limit + upper limit) / 2

Point estimate = (70+78)/2 = 74

Point estimate = $74

.

Margin of Error = (upper limit - lower limit) / 2

Margin of Error = (78-70) / 4

Margin of Error = $4

.

Standard error = Margin of error / z

z for 95% confidence interval = 1.96 ~ 2

Hence, Standard Error = 4/2 = 2

Thus standard error = 2

.

Statement 1: Since the sample size is very large, central limit theorem applies to this data, is definitely used while calculating the confidence interval. Thus this statement is TRUE

Statement 2: we are 95% confident, that average spending is between 70-78, with point estimate $74. This statement incorrectly mentions the point estimate to be $75, hence this statement is FALSE

Statement 3: Yes the margin of error is 4 , with standard error = 2, hence this statement is TRUE

Statement 4: In a 90% confidence interval the values will be as follows:

Z = 1.645 for 90% confidence level

SE = 2 [ it is the same for this data]

Hence Margin of error = Z * SE

Margin of error = 1.645 *2

Margin of error = 3.29

Thus this statement is FALSE.

Statement 5:

Standard error = Standard deviation/ square root of the sample size

we got standard error = 2

sample size = 436

Hence we can surely calculate the value of standard deviation with the given information

thus statement 5 is TRUE

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