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Summary STK110-TUTORIAL 9
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STK110 Preparation sheet: TUT 9 2023Question 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.
b. Use 𝛼 = 0.05 and determine the rejection rule using the critical value approach.
c. Calculate the value of the test statistic.
d. Make a decision.
e. Conclude and interpret.
f. Compute a 95% confidence interval for the population mean. Does it support your conclusion in e?
g. Compute the 𝑝-value.
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
b. Use 𝛼 = 0.01 and determine the rejection rule using the 𝑝-value approach.
c. Calculate the value of the test statistic
d. Make a decision
e. Conclude and interpret.
Question 3
A hamburger is considered “unhealthy” if the calorie content is more than 500. A health inspector noticed that
more than 60% of hamburgers are “unhealthy” and you want to investigate this observation. In a random
sample of 20 hamburgers, 16 were found to have a calorie content of more than 500.
Let: 𝑝 = the population proportion of “unhealthy” hamburgers
𝑝̅ = the sample proportion of “unhealthy” hamburgers.
a. Formulate the hypotheses
b. Use 𝛼 = 0.05 and determine the rejection rule using the critical value approach
c. Calculate the value of the test statistic
d. Make a decision
e. Conclude and interpret.
f. Compute the 𝑝-value.
Preliminary solutions:
Question 1 Question 2 Question 3
a. 𝐻0 : 𝜇 = 500; 𝐻𝑎 : a. 𝐻0 : 𝜇 ≥ 550; 𝐻𝑎 : a. 𝐻0 : ; 𝐻𝑎 : 𝑝 > 0.6
b. 𝑧 ≤ − or 𝑧 ≥ b. 𝑝 ≤ b. 𝑧 ≥
c. −2.0274 c. −1.911 c. 1.8257 (or 1.8265)
d. Reject/not 𝐻0 & reason d. Reject/not 𝐻0 & reason d. Reject/not 𝐻0 & reason
e. Interpretation e. Interpretation e. Interpretation
f. (460.6646; 499.3354) f. 0.0336
g. 0.0424
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