91Ó°ÊÓ

Nutritional information provided by Kentucky Fried Chicken (KFC) claims that each small bag of potato wedges contains 4.8 ounces of food and 280 calories. A sample of ten orders from KFC restaurants in New York and New Jersey averaged 358 calories. \(^{\star}\) a. If the sample standard deviation was \(s=54\), is there sufficient evidence to indicate that the average number of calories in small bags of KFC potato wedges is greater than advertised? Test at the \(1 \%\) level of significance. b. Construct a \(99 \%\) lower confidence bound for the true mean number of calories in small bags of KFC potato wedges. c. On the basis of the bound you obtained in part (b), what would you conclude about the claim that the mean number of calories exceeds \(280 ?\) How does your conclusion here compare with your conclusion in part (a) where you conducted a formal test of hypothesis?

Step by step solution

01

Define the Hypotheses

We want to test whether the average number of calories is greater than 280. The null hypothesis (H_0) is that the mean is 280, and the alternative hypothesis (H_1) is that the mean is greater than 280:\[H_0: \mu = 280\]\[H_1: \mu > 280\]

02

Determine the Significance Level and Critical Value

The level of significance is given as 1%, meaning we will reject the null hypothesis if the p-value is less than 0.01. We use a t-distribution with 9 degrees of freedom (n-1 = 10-1 = 9). The critical t-value for a one-tailed test at 0.01 significance level can be found using a t-table or calculator.

03

Calculate the Test Statistic

We use the formula for the t-statistic:\[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\]where \(\bar{x} = 358\), \(\mu_0 = 280\), \(s = 54\), and \(n = 10\). Plugging in these values gives:\[t = \frac{358 - 280}{54/\sqrt{10}}\approx 4.845\]

04

Make a Decision Based on the Test Statistic

Compare the calculated t-statistic to the critical value obtained for 9 degrees of freedom at 1% significance (approximately 2.821 from the t-table). Since 4.845 > 2.821, we reject the null hypothesis. This provides evidence that the average number of calories is greater than advertised.

05

Construct a 99% Lower Confidence Bound

To create a 99% lower confidence bound for the true mean calories, we use:\[\bar{x} - t_{0.01} \cdot \frac{s}{\sqrt{n}}\].Using the same \(t_{0.01} = 2.821\) and our values:\[LCB = 358 - 2.821 \cdot \frac{54}{\sqrt{10}} \approx 314.725\]

06

Interpret the Lower Confidence Bound

The lower confidence bound is approximately 314.725 calories, which suggests that the true mean is higher than 280 calories, lending support to the earlier hypothesis test conclusion.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval

A confidence interval provides a range, calculated from the data, that is likely to include the true population parameter. In this exercise, we are interested in the mean number of calories in KFC potato wedges.
To construct a confidence interval, we used the sample mean and the sample standard deviation. By doing this, we aim to have a range where the true mean falls into it, with a certain degree of certainty.

• A 99% confidence level means there's a 99% likelihood the interval contains the true mean.
• It's calculated as: \( \bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} \).

In this case, we found the lower bound of the confidence interval to be approximately 314.725 calories.
When the lower bound of a confidence interval is greater than the claimed 280 calories, this indicates that the true mean calorie count is likely higher than 280.

t-distribution

The t-distribution is vital for hypothesis testing, especially when dealing with small sample sizes. It looks similar to the standard normal distribution but has heavier tails, which account for the variability in smaller samples.

• It's used when the sample size is small (n < 30) and/or the population standard deviation is unknown.
• The shape of the t-distribution depends on the "degrees of freedom," which is equal to the sample size minus one (n-1).

In our exercise, we used a t-distribution with 9 degrees of freedom because we had 10 samples.
The t-value we found was crucial in determining if the average calorie count was greater than the advertised amount.

Significance Level

The significance level in hypothesis testing measures how willing we are to make a type I error, which occurs when we reject a true null hypothesis.
In our scenario, the significance level was set to 1% (0.01).

• This low percentage indicates strong evidence is required to reject the null hypothesis.
• A significance level of 0.01 means there's a 1% probability of rejecting the null if it is indeed true.

It acts as a threshold, and if our p-value or test statistic surpasses it, we find sufficient evidence against the null hypothesis.
In the exercise, our test statistic was well above the critical value for the 1% significance level, leading us to reject the claim that the mean is 280 calories.

Sample Standard Deviation

The sample standard deviation tells us how much variation exists in our sample data. It's a measure of the dispersion or spread of the sample values.
It is denoted by \(s\) and is crucial in calculating both the t-statistic and the confidence interval.

• A larger standard deviation indicates more variability in the data.
• It directly affects the width of the confidence interval and scale of the test statistic.

In the provided step-by-step solution, the sample standard deviation was 54, showing a reasonable spread in the calorie counts among the sampled orders.
This value was integral to our calculations and conclusions about the mean calorie count.