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Unlike most packaged food products, alcohol beverage container labels are not required to show calorie or nutrient content. The article 鈥淲hat Am I Drinking? The Effects of Serving Facts Information on Alcohol Beverage Containers鈥 (J. of Consumer Affairs, 2008: 81鈥99) reported on a pilot study in which each of 58 individuals in a sample was asked to estimate the calorie content of a 12-oz can of beer known to contain 153 calories. The resulting sample mean estimated calorie level was 191 and the sample standard deviation was 89. Does this data suggest that the true average estimated calorie content in the population sampled exceeds the actual content? Test the appropriate hypotheses at significance level .001.

Short Answer

Expert verified

There is sufficient evidence to support the claim that the true average estimated calorie content in the population sampled exceed the actual content.

Step by step solution

01

Step 1:Null hypothesis.

The null hypothesis, denoted by H0, is the claim that is initially assumed to be true (the 鈥減rior belief鈥 claim). The alternative hypothesis, denoted by Ha, is the assertion that is contradictory to H0.

The null hypothesis will be rejected in favour of the alternative hypothesis only if sample evidence suggests that H0 is false. If the sample does not strongly contradict H0, we will continue to believe in the plausibility of the null hypothesis. The two possible conclusions from a hypothesis-testing analysis are then reject H0 or fail to reject H0.

02

To find the z-value.

Given claim: Average exceed the actual content of \(153\) calories.

The null hypothesis states that the population mean is equal to the value mentioned in the claim:

\({H_0}:\mu = 153\)

The alternative hypothesis states the claim:

\({H_a}:\mu > 153\)

The sampling distribution of the sample mean \(\overline x \) has mean \(\mu \)and standard deviation \(\frac{\sigma }{{\sqrt n }}\).

The z-core is the value decreased by the mean, divided by the standard deviation:

\(z = \frac{{\overline x - \mu }}{{\sigma /\sqrt n }}\)

\(\begin{array}{l}z = \frac{{191 - 153}}{{89/\sqrt {58} }}\\z \approx 3.25\end{array}\)

03

Hypothesis is reject or not.

The P-value is the probability of obtaining a value more extreme or equal to the standard deviation test statistic z, assuming that the null hypothesis is true. Determine the probability table in the appendix.

\(\begin{array}{l}P = P(Z > 3.25)\\ = 1 - p(z < 3.25)\\ = 1 - 0.9994\\ = 0.0006\end{array}\]

The \(P\]-value is smaller than the significance level \(\alpha \],then the null hypothesis is rejected.

\(P < 0.001\]\( \Rightarrow \]Reject \({H_0}\]

There is sufficient evidence to support the claim that the true average estimated calorie content in the population sampled exceed the actual content.

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