CEGEP CHAMPLAIN - ST. LAWRENCE
201-510-LW: Business Statistics
Patrice Camiré
Problem Sheet #16
Hypothesis Testing - One Sample
1. A small brewery has purchased a used automatic bottle filler. The technical specifications state that
the volume distributed in a bottle is normally distributed with a mean of 541 ml and a standard
deviation of 2 ml. To test whether or not the bottle filler is still calibrated, the brewery decides to fill
20 bottles and obtains a sample mean of 540 ml and a sample standard deviation of 2.6 ml.
(a) Test that the specifications for the standard deviation are correct at the 5% level of significance.
(b) Would your conclusion in part (a) be different if the level of significance was changed to 2.5%?
Justify.
(c) Test the claim that the mean is different than 541 ml at the 5% level of significance.
(d) Test the claim that the mean is less than 541 ml at the 5% level of significance.
(e) What if a sample mean of 539.4 ml was obtained? Formulate a proper hypothesis about the mean
and test it at the 1% level of significance.
2. A company manufactures and sells cases of light bulbs. They claim that the number of defective light
bulbs in a case is normally distributed with a mean of 5 light bulbs. Your office purchases 8 cases.
Here is the number of defective light bulbs in each of the cases:
4
3
6
6
7
5
8
6
(a) Test, at the 5% level of significance, the claim that the mean number of defective light bulbs in
a case is greater than 5.
(b) Construct a 90% confidence interval for the true mean number of defective light bulbs in a case.
(c) Is your confidence interval in part (b) consistent with your conclusion in part (a)? Justify.
(d) Test the claim that the standard deviation is less than 2 at the 10% level of significance.
(e) Is there a way to change the significance level to obtain a different conclusion in part (d)? Justify.
3. A furniture manufacturer has been awarded a contract to manufacture the chairs used in classrooms of
several high schools in the province. The specifications of the chairs are based mainly on the average
height of a student. Kevin, who is in charge of the design, found a statistical study from 1990 stating
that the average height of a high school student is 5.67 feet. Kevin does not believe this number to still
be accurate and decide to test this claim by randomly sampling 80 high school students; he obtained
a sample mean of 5.83 feet and a sample standard deviation of 0.41 feet.
(a) Test, at the 1% level of significance, the claim that the current average height of a high school
student is different from the one in 1990.
(b) Test, at the 1% level of significance, the claim that the current average height of a high school
student is greater than 5.67.
(c) Construct a 90% confidence interval for the true average height of a high school student.
(d) Is your confidence interval in part (c) consistent with your conclusion in part (b)? Justify.
(e) Test the claim that the standard deviation is greater than 0.35 at the 2.5% level of significance.
(f) Is the conclusion in part (e) different if the level of significance is changed to 1%? Justify.
4. A phone company determined 10 years ago that the length of a phone conversation was normally
distributed with a mean of 9 minutes. Today, the phone company believes that people talk more
often on the phone, but have on average shorter conversations. The company recorded the length
of 23 random phone conversations and obtained a sample mean of 7 minutes and a sample standard
deviation of 3 minutes.
(a) Test, at the 1% level of significance, the claim that phone conversations are shorter today than
they were 10 years ago.
(b) Construct a 99% confidence interval for the true average length of a phone conversation.
(c) Is your confidence interval in part (b) consistent with your conclusion in part (a)? Justify.
(d) Test the claim that the standard deviation is different than 4 at the 5% level of significance.
(e) Test the claim that the standard deviation is less than 4 at the 5% level of significance.
5. A study showed that 5 years ago, people driving to work in Quebec city needed on average 19.2 minutes
in the morning. To determine if this has changed, you decide to randomly sample 36 people who drive
to get to work and obtain a mean of 21.7 minutes and a standard deviation of 5.1 minutes.
(a) Test, at the 5% level of significance, the claim that it takes people longer to drive to work in the
morning today than it did 5 years ago.
(b) Construct a 95% confidence interval for today’s true mean time to drive to work in the morning.
(c) Is your confidence interval in part (b) consistent with your conclusion in part (a)? Justify.
(d) Test the claim that the standard deviation is greater than 4.7 at the 5% level of significance.
6. A friend of yours claims to be clairvoyant. To test this claim, you roll a die 100 times and ask your
friend to predict the figure on the upper face each time. Your friend was able to predict the figure 25
times out of 100.
(a) Test, at the 1% level of significance, the claim that your friend is clairvoyant.
(b) What if you had used a 2% level of significance? Would your conclusion be different?
7. Following the controversy from the first experiment, your friend is still convinced that she is clairvoyant,
so you decide to design a new experiment. You draw a card from a deck of 52 cards and your friend
has to predict the suit of the drawn card. You repeat this 50 times and note that your friend was right
14 times.
(a) Test, at the 3% level of significance, the claim that your friend is clairvoyant.
(b) This time, is there a way to change the level of significance so as to reach a different conclusion?
Justify.
8. Raymond is a lawyer for a private law firm advertising that their lawyers win 76% of all their cases.
Over the past year, Raymond was in charge of 36 cases and won 21. Raymond was just fired for being
a sub par employee.
(a) Was Raymond’s firing justified? Test the claim that Raymond is a sub par employee at the 1%
level of significance.
(b) Is there any way to adjust the level of significance so as not to fire Raymond? Justify.
(c) Construct a 95% confidence interval for Raymond’s true proportion of winning cases. Is this
consistent with your conclusion in part (a)? Justify.
9. A biologist has produced a new variety of tomato plants that is less sensitive to cool temperatures.
The biologist knows that for the original tomato plants, the proportion reaching maturity is 87%.
This proportion is still unknown for the new variety, but the biologist claims that it is the same as the
original variety. To test this claim, 300 seeds from the new variety are planted and it is later found
that 257 reached maturity.
(a) Using a 4% level of significance, test the claim that the proportion of the new variety of tomato
plants reaching maturity is different than the original variety.
(b) Can you adjust α to obtain a different conclusion? Justify.
(c) Construct a 98% confidence interval for the true proportion of the new variety of tomato plants
reaching maturity. Is this consistent with your conclusion in part (a)? Justify.
Answers
1. (a) H0 : σ = 2, H1 : σ > 2, α = 0.05, df = 19, χ2obs ≈ 32.11, R = [30.14, ∞).
We reject H0 .
(b) α = 0.025, R = [32.85, ∞).
Yes, we now accept H0 .
(c) H0 : µ = 541, H1 : µ 6= 541, α = 0.05, df = 19, tobs ≈ −1.72, R = (−∞, −2.093] ∪ [2.093, ∞).
We accept H0 .
(d) H0 : µ = 541, H1 : µ < 541, α = 0.05, df = 19, tobs ≈ −1.72, R = (−∞, −1.729].
We accept H0 .
(e) H0 : µ = 541, H1 : µ < 541, α = 0.01, df = 19, tobs ≈ −2.75, R = (−∞, −2.539].
We reject H0 .
2. x = 5.625, s2 = 17.875/7, s ≈ 1.598.
(a) H0 : µ = 5, H1 : µ > 5, α = 0.05, df = 7, tobs ≈ 1.106, R = [1.895, ∞).
We accept H0 .
(b) 4.55 ≤ µ ≤ 6.70
(c) Yes, since µ = 5 is contained in our confidence interval.
(d) H0 : σ = 2, H1 : σ < 2, α = 0.1, df = 7, χ2obs ≈ 4.469, R = [0, 2.83].
We accept H0 .
(e) No, since we would need α > 25%, which is too large.
3. (a) H0 : µ = 5.67, H1 : µ 6= 5.67, α = 0.01, df = 79, tobs ≈ 3.49, R = (−∞, −2.632] ∪ [2.632, ∞).
We reject H0 .
(b) H0 : µ = 5.67, H1 : µ > 5.67, α = 0.01, df = 79, tobs ≈ 3.49, R = [2.368, ∞).
We reject H0 .
(c) 5.754 ≤ µ ≤ 5.906
(d) Yes, since every value in our confidence interval is greater than 5.67.
(e) H0 : σ = 0.35, H1 : σ > 0.35, α = 0.025, df = 79, χ2obs ≈ 108.41, R = [106.6, ∞).
We reject H0 .
(f) Yes, since R = [112.3, ∞), so we accept H0 .
4. (a) H0 : µ = 9, H1 : µ < 9, α = 0.01, df = 22, tobs ≈ −3.20, R = (−∞, −2.508]
We reject H0 .
(b) 5.24 ≤ µ ≤ 8.76
(c) Yes, since all the values in our confidence interval are less than 9.
(d) H0 : σ = 4, H1 : σ 6= 4, α = 0.05, df = 22, χ2obs ≈ 12.375, R = [0, 10.98] ∪ [36.78, ∞)
We accept H0 .
(e) H0 : σ = 4, H1 : σ < 4, α = 0.05, df = 22, χ2obs ≈ 12.375, R = [0, 12.34]
We accept H0 .
5. (a) H0 : µ = 19.2, H1 : µ > 19.2, α = 0.05, df = 35, tobs ≈ 2.94, R = [1.69, ∞).
We reject H0 .
(b) 19.97 ≤ µ ≤ 23.43
(c) Yes, since all the values in our confidence interval are greater than 19.2.
(d) H0 : σ = 4.7, H1 : σ > 4.7, α = 0.05, df = 35, χ2obs ≈ 41.211.
If we use df = 30, then R = [43.77, ∞).
If we use df = 40, then R = [55.76, ∞).
In either case, we accept H0 .
6. (a) H0 : p = 1/6, H1 : p > 1/6, α = 0.01, q = 5/6, n = 100, r = 25, p̂ = 1/4, np > 5, nq > 5,
zobs ≈ 2.24, R = [2.33, ∞).
We accept H0 .
(b) Yes, since R = [2.05, ∞), we would reject H0 .
7. (a) H0 : p = 1/4, H1 : p > 1/4, α = 0.03, q = 3/4, n = 50, r = 14, p̂ = 14/50, np > 5, nq > 5,
zobs ≈ 0.49, R = [1.88, ∞).
We accept H0 .
(b) No. To reject H0 , we would need α ≥ 31.21%, which is too large.
8. (a) H0 : p = 0.76, H1 : p < 0.76, α = 0.01, q = 0.24, n = 36, r = 21, p̂ = 21/36, np > 5, nq > 5,
zobs ≈ −2.48, R = (−∞, −2.33].
We reject H0 .
(b) No. To accept H0 , we would need α < 0.66%, which is too small.
(c) 0.422 ≤ p ≤ 0.744
Our confidence interval is consistent with our conclusion in part (a), since all the values in our
confidence interval are below 0.76.
9. (a) H0 : p = 0.87, H1 : p 6= 0.87, α = 0.04, q = 0.13, n = 300, r = 257, p̂ = 257/300, np > 5, nq > 5,
zobs ≈ −0.687, R = (−∞, −2.05] ∪ [2.05, ∞).
We accept H0 .
(b) No. To reject H0 , we would need α ≥ 49%, which is too large.
(c) 0.810 ≤ p ≤ 0.904
Our confidence interval is consistent with our conclusion in part (a), since it contains p = 0.87.