The file Danish Coffee
Download Danish Coffee
contains a random sample of 144 Danish coffee drinkers and measures the annual coffee consumption in kilograms for each sampled coffee drinker. A marketing research firm wants to use this information to develop an advertising campaign to increase Danish coffee consumption.
Develop and interpret a 90% confidence interval estimate for the mean annual coffee consumption of Danish coffee drinkers.

Question

The file Danish Coffee 
Download Danish Coffee
contains a random sample of 144 Danish coffee drinkers and measures the annual coffee consumption in kilograms for each sampled coffee drinker. A marketing research firm wants to use this information to develop an advertising campaign to increase Danish coffee consumption.
Develop and interpret a 90% confidence interval estimate for the mean annual coffee consumption of Danish coffee drinkers.

BlackTom AI · Accepted Answer

The question asks us to develop and interpret a 90% confidence interval for the mean annual coffee consumption from a random sample of 144 Danish coffee drinkers. The core idea is that a 90% CI for a mean uses the sample mean plus or minus a margin of error based on the standard error and the critical t (or z) value for n-1 degrees of freedom. Now let's evaluate each answer option in light of that principle.

Option A: (6.4257, 6.6479)
This interval would imply a relatively small margin of error around the center. If the sample mean were near 6.5368 (the midpoint), the half-width would be about 0.1111. Whether this is consistent with a 90% CI depends on the sample standard deviation and the t-value. Given n = 144, the t-value for 89 degrees of freedom at 90% two-sided is approximately 1.66 to 1.66, so the standard error would need to be around 0.066 or so to yield a half-width near 0.111. Without the actual s value, we can’t confirm, but this interval’s width might be plausible if the sample variability was modest. However, we should compare to other options for context.

Option B: (6.3881, 6.6855)
This interval is centered at about 6.5363 with a half-width of roughly 0.1247. A 90% CI with n = 144 would use the t critical value near 1.66 and SE = s/√n. To achieve a half-width of 0.1247, the estimated standard error would be about 0.1247 × (1/1.66) ≈ 0.075. Since SE = s/√n and √n = 12, this implies s ≈ 0.9. That is a plausible sample standard deviation for annual coffee consumption in kilograms with 144 observations, so this interval is plausible and consistent with a reasonable s. It seems like a reasonable CI for the mean.

Option C: (6.3366, 6.7370)
This interval has a center around 6.5378 with a half-width of about 0.2002, which is noticeably wider than the previous options. A 90% CI of this width would require an SE around 0.120 as well (since 0.2002 ≈ 1.66 × SE). That would imply s ≈ SE × √n ≈ 0.120 × 12 ≈ 1.44, which is quite large for annual coffee consumption variability. While not impossible, this would indicate a much more variable sample than typical, making this wider interval less likely but still possible if the data were highly dispersed.

Option D: (6.1768, 6.8968)
This interval has a center around 6.5368 with a half-width of 0.3599, which is quite wide. The corresponding SE would be roughly 0.3599 / 1.66 ≈ 0.216. That leads to s ≈ 0.216 × 12 ≈ 2.59, which would indicate extremely high variability in annual coffee consumption among the 144 drinkers. Such a wide interval is possible only if the data were highly dispersed or if there were outliers, making this the least likely CI for a typical sample of this nature, though still mathematically valid.

In summary, each option reflects a 90% confidence interval that could arise from different sample standard deviations and means. The most plausible interval given a reasonable level of variability in coffee consumption is typically the one with a moderate half-width around 0.12 to 0.13, which aligns with Option B (6.3881, 6.6855). Options A and C have narrower or wider widths that imply smaller or larger SE values, and Option D implies a particularly large variability that is less plausible without extreme data features.

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Part 7 of 8 The 95% confidence interval of for the annual return is ( [ Select ] -0.0536 -0.0012 -0.028 , [ Select ] 0.0792 0.0524 0.0536 )

Part 8 of 8 The 95% confidence interval of for the two year period return is ( -0.0239 , 0.0898 )

Part 3 of 3 The 95% Confidence Interval is ( -0.2158 , 0.2186 )

Part 8 of 8 The 95% confidence interval of for the two year period return is ( [ Select ] -0.0239 0.0140 -0.0758 , [ Select ] 0.0898 0.1277 0.0758 )

Part 7 of 8 The 95% confidence interval of for the annual return is ( [ Select ] -0.028 -0.0012 -0.0536 , [ Select ] 0.0536 0.0524 0.0792 )

Part 8 of 8 The 95% confidence interval of for the two year period return is ( [ Select ] 0.0140 -0.0758 -0.0239 , [ Select ] 0.1277 0.0758 0.0898 )

Part 7 of 8 The 95% confidence interval of for the annual return is ( [ Select ] -0.0536 -0.028 -0.0012 , [ Select ] 0.0792 0.0536 0.0524 )

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