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Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. The average 1 -ounce chocolate chip cookie contains 110 calories. A random sample of 15 different brands of 1 -ounce chocolate chip cookies resulted in the following calorie amounts. At the \(\alpha=0.01\) level, is there sufficient evidence that the average calorie content is greater than 110 calories? $$ \begin{array}{lllllllll} 100 & 125 & 150 & 160 & 185 & 125 & 155 & 145 & 160 \\ 100 & 150 & 140 & 135 & 120 & 110 & & & \end{array} $$

Short Answer

Expert verified
Yes, the evidence supports the claim that the average calorie content is greater than 110 calories.

Step by step solution

01

State the hypotheses and identify the claim

To perform hypothesis testing, we need to state the null hypothesis and the alternative hypothesis. Here, our hypotheses are as follows: - Null hypothesis \( H_0 \): The average calorie content of 1-ounce chocolate chip cookies is 110 calories. Mathematically, \( \mu = 110 \).- Alternative hypothesis \( H_1 \): The average calorie content is greater than 110 calories. Mathematically, \( \mu > 110 \). This is a right-tailed test since we are testing if the mean is greater than a specific value. The claim is that the average calorie content is greater than 110 calories.
02

Find the critical value(s)

Since the sample size is small (n < 30), we use the t-distribution to determine the critical value. The significance level given is \( \alpha = 0.01 \) for a one-tailed test. We have \( n - 1 = 15 - 1 = 14 \) degrees of freedom. Using a t-distribution table, the critical t-value at \( \alpha = 0.01 \) and 14 degrees of freedom is \( t_{\alpha} = 2.624 \).
03

Find the test value

First, calculate the sample mean \( \bar{x} \) and the sample standard deviation \( s \) from the given data:\[\bar{x} = \frac{100 + 125 + 150 + 160 + 185 + 125 + 155 + 145 + 160 + 100 + 150 + 140 + 135 + 120 + 110}{15} \approx 137.33\]Next, calculate the sample standard deviation using the formula:\[s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \approx 25.45\]Then, calculate the test statistic using the t-formula:\[t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{137.33 - 110}{25.45/\sqrt{15}} \approx 3.51\]
04

Make the decision

Compare the test statistic \( t = 3.51 \) with the critical value \( t_{\alpha} = 2.624 \). Since the test statistic is greater than the critical value, we reject the null hypothesis.
05

Summarize the results

We have sufficient evidence at the \( \alpha=0.01 \) level to support the claim that the average calorie content of the 1-ounce chocolate chip cookies is greater than 110 calories.

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Key Concepts

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

Understanding Critical Value
In hypothesis testing, the critical value is a key component to determine whether to reject the null hypothesis. It's a threshold set based on the significance level \(\alpha\) and the distribution being used. When we perform the test, we compare the calculated test statistic against this critical value.

If the test statistic exceeds the critical value, we reject the null hypothesis. This means our sample provides enough evidence to support the alternative hypothesis.

In our problem, the critical value of 2.624 was obtained from the t-distribution table. This critical value is specifically for a one-tailed test with 14 degrees of freedom at the 1% significance level. It's important to choose the correct distribution and degrees of freedom to get the right critical value.
  • For a higher significance level, the critical value would be smaller.
  • Pushing beyond the critical value increases the confidence in rejecting the null hypothesis.
The T-Distribution
The t-distribution is an essential concept in statistics, especially useful when dealing with small sample sizes (typically less than 30). It's quite similar to the standard normal distribution but has heavier tails, which accounts for the variability in smaller samples.

In hypothesis testing, we often use the t-distribution when the population standard deviation is unknown. This distribution provides a range of potential values that the test statistic can take, based on the sample size and the desired level of confidence.
  • The shape of the t-distribution depends on the degrees of freedom, \(n-1\), where \(n\) is the sample size.
  • With larger samples, the t-distribution closely resembles a normal distribution.

For our example, with n=15, we have 14 degrees of freedom, which results in selecting the critical value from the t-distribution table accurately. Using this approach ensures the test results are reliable and accounts for sample variability.
Sample Mean in Context
The sample mean is a crucial component in hypothesis testing as it represents the average of the sample data. It's essentially the best estimate we have of the population mean when given a sample.

In our specific scenario, after calculating the sum of all calorie contents from the sample data, we obtained a sample mean \(\bar{x}\approx137.33\) calories. This mean is then used as part of the formula for the test statistic.

Understanding the sample mean helps set a reference point for our expectations. We compare this with the hypothesized mean to understand deviations within the sample data accurately. Small differences generally imply the null hypothesis might still hold, while larger deviations suggest otherwise. This process builds towards supporting or rejecting the initial hypothesis.
  • Always calculate the sample mean accurately for precise results.
  • A sample mean significantly different from the hypothesized mean could indicate a need to reject the null hypothesis.
Understanding the Test Statistic
The test statistic is a standardized value that helps determine the degree of deviation from the null hypothesis within a given sample. It measures how far our sample mean is from the hypothesized population mean, scaled by the sample standard deviation.

In our exercise, we calculated the test statistic using the formula: \[t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \] Plugging in the numbers, we found that the test statistic is approximately 3.51.

This value indicates how "extreme" the situation is compared to our null hypothesis. For our specific case, a test statistic of 3.51 is significant enough to exceed the critical value of 2.624, prompting us to reject the null hypothesis.
  • The larger the test statistic, the further the sample mean is from the hypothesized mean.
  • This high difference generally leads to rejecting the null hypothesis in favor of the alternative hypothesis.
Ensuring the proper calculation of the test statistic is vital for drawing accurate and meaningful conclusions from your hypothesis test.

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Most popular questions from this chapter

Assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Daily weather observations for southwestern Pennsylvania for the first three weeks of January for randomly selected years show daily high temperatures as follows: \(55,44,51,59,62,\) \(60,46,51,37,30,46,51,53,57,57,39,28,37,35,\) and 28 degrees Fahrenheit. The normal standard deviation in high temperatures for this time period is usually no more than 8 degrees. A meteorologist believes that with the unusual trend in temperatures the standard deviation is greater. At \(\alpha=0.05,\) can we conclude that the standard deviation is greater than 8 degrees?

Workers with a formal arrangement with their employer to be paid for time worked at home worked an average of 19 hours per week. A random sample of 15 mortgage brokers indicated that they worked a mean of 21.3 hours per week at home with a standard deviation of 6.5 hours. At \(\alpha=0.05,\) is there sufficient evidence to conclude a difference? Construct a \(95 \%\) confidence interval for the true mean number of paid working hours at home. Compare the results of your confidence interval to the conclusion of your hypothesis test and discuss the implications.

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. In the Journal of the American Dietetic Association, it was reported that \(54 \%\) of kids said that they had a snack after school. A random sample of 60 kids was selected, and 36 said that they had a snack after school. Use \(\alpha=0.01\) and the \(P\) -value method to test the claim. On the basis of the results, should parents be concerned about their children eating a healthy snack?

Many people believe that the average number of Facebook friends is \(338 .\) The population standard deviation is 43.2 . A random sample of 50 high school students in a particular county revealed that the average number of Facebook friends was \(350 .\) At \(\alpha=0.05,\) is there sufficient evidence to conclude that the mean number of friends is greater than \(338 ?\)

For each conjecture, state the null and alternative hypotheses. a. The average age of first-year medical school students is at least 27 years. b. The average experience (in seasons) for an NBA player is 4.71 c. The average number of monthly visits/sessions on the Internet by a person at home has increased from 36 in 2009 d. The average cost of a cell phone is \(\$ 79.95 .\) e. The average weight loss for a sample of people who exercise 30 minutes per day for 6 weeks is 8.2 pounds.

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