STK110 TUT Preparation sheet: memo 2023
Question 1
The calorie content of hamburgers is normally distributed with mean 500 and standard deviation of 39.46.
A health inspector wants to test whether the mean calorie content of hamburgers is different from 500. A
random sample of 16 hamburgers has an average calorie content of 480.
Let: 𝜇 = the population mean calorie content of hamburgers
𝑥̅ = the sample mean calorie content of a random sample of 16 hamburgers
a. Formulate the hypotheses.
𝐻 :𝜇 500
𝐻 :𝜇 500
b. Use 𝛼 0.05 and determine the rejection rule using the critical value approach.
Reject 𝐻 if 𝑧 1.96 or 𝑧 1.96
c. Calculate the value of the test statistic.
𝑥 𝜇 480 500
𝑧 𝜎 2.0274
39.46
√𝑛 √16
d. Make a decision.
2.0274 1.96, therefore Reject 𝐻 at the 0.05 level of significance.
e. Conclude and interpret.
At a 0.05 level of significance there is sufficient evidence to conclude that the mean calorie content of
hamburgers is not 500.
f. Compute a 95% confidence interval for the population mean. Does it support your conclusion in e?
𝜎 39.46
𝑥̅ 𝑧 480 1.96 480 19.3354 460.6646; 499.3354
√𝑛 √16
It supports the conclusion in e, because 500 does not fall into the interval - Reject 𝐻 at the 0.05 level
of significance.
g. Compute the 𝑝-value.
𝑝 2𝑃 𝑧 2.03 2 0.0212 0.0424
Question 2
At a certain fast-food restaurant 15 randomly chosen hamburgers were tested for their calorie content. The
manager claims that the mean calorie content of hamburgers is at least 550.The results are:
420 450 450 450 450 480 550 555 560 560 570 575 575 575 580
Assume: Data is normally distributed.
Let: 𝜇 = the population mean calorie content.
Given: 𝑥̅ 520; 𝑠 60.8
a. Formulate the hypotheses.
𝐻 :𝜇 550
𝐻 :𝜇 550
b. Use 𝛼 0.01 and determine the rejection rule using the 𝑝-value approach.
Reject 𝐻 if 𝑝 0.01
c. Calculate the value of the test statistic.
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Question 1
The calorie content of hamburgers is normally distributed with mean 500 and standard deviation of 39.46.
A health inspector wants to test whether the mean calorie content of hamburgers is different from 500. A
random sample of 16 hamburgers has an average calorie content of 480.
Let: 𝜇 = the population mean calorie content of hamburgers
𝑥̅ = the sample mean calorie content of a random sample of 16 hamburgers
a. Formulate the hypotheses.
𝐻 :𝜇 500
𝐻 :𝜇 500
b. Use 𝛼 0.05 and determine the rejection rule using the critical value approach.
Reject 𝐻 if 𝑧 1.96 or 𝑧 1.96
c. Calculate the value of the test statistic.
𝑥 𝜇 480 500
𝑧 𝜎 2.0274
39.46
√𝑛 √16
d. Make a decision.
2.0274 1.96, therefore Reject 𝐻 at the 0.05 level of significance.
e. Conclude and interpret.
At a 0.05 level of significance there is sufficient evidence to conclude that the mean calorie content of
hamburgers is not 500.
f. Compute a 95% confidence interval for the population mean. Does it support your conclusion in e?
𝜎 39.46
𝑥̅ 𝑧 480 1.96 480 19.3354 460.6646; 499.3354
√𝑛 √16
It supports the conclusion in e, because 500 does not fall into the interval - Reject 𝐻 at the 0.05 level
of significance.
g. Compute the 𝑝-value.
𝑝 2𝑃 𝑧 2.03 2 0.0212 0.0424
Question 2
At a certain fast-food restaurant 15 randomly chosen hamburgers were tested for their calorie content. The
manager claims that the mean calorie content of hamburgers is at least 550.The results are:
420 450 450 450 450 480 550 555 560 560 570 575 575 575 580
Assume: Data is normally distributed.
Let: 𝜇 = the population mean calorie content.
Given: 𝑥̅ 520; 𝑠 60.8
a. Formulate the hypotheses.
𝐻 :𝜇 550
𝐻 :𝜇 550
b. Use 𝛼 0.01 and determine the rejection rule using the 𝑝-value approach.
Reject 𝐻 if 𝑝 0.01
c. Calculate the value of the test statistic.
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