Calculative Questions on Hypothesis Testing

Question 1:
A manufacturing company claims that their light bulbs last for an average of 1500 hours with a standard deviation of 120 hours. A consumer group tests 100 bulbs and finds an average life of 1470 hours. Test the manufacturer's claim at a 5% level of significance.

Solution:

H₀: μ = 1500 hours
H₁: μ ≠ 1500 hours
Given: μ₀ = 1500, σ = 120, n = 100, x̄ = 1470, α = 0.05
Test statistic: $Z = \frac{(\bar{x} - μ_{0})}{σ / \sqrt{n}} = \frac{(1470 - 1500)}{120 / \sqrt{100}} = \frac{- 30}{120 / 10} = \frac{- 30}{12} = - 2.5$
Critical value for α = 0.05 (two-tailed): $Z_{(α / 2)} = \pm 1.96$
Since $| - 2.5 | > 1.96$ , we reject H₀
Conclusion: There is sufficient evidence to conclude that the mean life of the bulbs is not 1500 hours.

Question 2:
The mean breaking strength of a certain type of rope is claimed to be 80 kg with a standard deviation of 5 kg. A random sample of 36 pieces of rope has a mean breaking strength of 78.5 kg. At a 1% level of significance, is there evidence that the mean breaking strength is less than claimed?

Solution:

H₀: $μ = 80 kg$
H₁: $μ < 80 kg$ (one-tailed)
Given: $μ 0 = 80$ , $σ = 5$ , $n = 36$ , $\bar{x} = 78.5$ , $α = 0.01$
Test statistic: $Z = \frac{\bar{x} - μ_{0}}{σ / \sqrt{n}} = \frac{78.5 - 80}{5 / \sqrt{36}} = \frac{- 1.5}{0.833} = - 1.8$
Critical value for $α = 0.01$ (one-tailed): $Z_{α} = - 2.33$
Since $Z = - 1.8 > - 2.33$ , we fail to reject H₀
Conclusion: There is insufficient evidence to conclude that the mean breaking strength is less than 80 kg at the 1% significance level.

Question 3:
Two different manufacturing processes are used to produce metal rods. Process A produces rods with a mean length of 12 cm and standard deviation of 0.5 cm. Process B produces rods with a mean length of 12.3 cm and standard deviation of 0.6 cm. If 50 rods are randomly selected from process A and 60 rods from process B, test at 5% significance level whether the mean lengths are different.

Solution:

H₀: $μ ₁ = μ ₂$ (or $μ ₁ - μ ₂ = 0$ )
H₁: $μ ₁ \neq μ ₂$ (or $μ ₁ - μ ₂ \neq 0$ )
Given: $μ ₁ = 12$ , $σ ₁ = 0.5$ , $n ₁ = 50$ , $μ ₂ = 12.3$ , $σ ₂ = 0.6$ , $n ₂ = 60$ , $α = 0.05$
Test statistic: $Z = \frac{μ ₁ - μ ₂ - 0}{\sqrt{\frac{σ^{₁}}{2} + \frac{σ^{₂}}{2}}} = \frac{-0.3}{\sqrt{0.005 + 0.006}} = \frac{-0.3}{\sqrt{0.011}} = \frac{-0.3}{0.105} = -2.86$
Critical value for $α = 0.05$ (two-tailed): $Z_{α / 2} = \pm 1.96$
Since $| -2.86 | > 1.96$ , we reject H₀
Conclusion: There is significant evidence that the mean lengths of the rods produced by the two processes are different.

Question 4:
A sample of 16 computer processors had a mean processing speed of 3.2 GHz with a sample standard deviation of 0.4 GHz. Test whether the population mean processing speed is different from 3.0 GHz at a 5% level of significance.

Solution:

H₀: μ = 3.0 GHz
H₁: μ ≠ 3.0 GHz
Given: x̄ = 3.2, s = 0.4, n = 16, μ₀ = 3.0, α = 0.05
Test statistic: $t = \frac{x̄ - μ_{0}}{s / \sqrt{n}} = \frac{3.2 - 3.0}{0.4 / \sqrt{16}} = \frac{0.2}{0.4 / 4} = \frac{0.2}{0.1} = 2.0$
Degrees of freedom = n - 1 = 16 - 1 = 15
Critical value for α = 0.05 (two-tailed) with df = 15: $t_{α / 2} = \pm 2.131$
Since |2.0| < 2.131, we fail to reject H₀
Conclusion: There is insufficient evidence to conclude that the mean processing speed is different from 3.0 GHz.

Question 5:
A new teaching method is claimed to improve test scores. A random sample of 12 students who were taught using this method had a mean score of 76 with a sample standard deviation of 8. The mean score with the traditional method is 70. At a 1% significance level, is there evidence that the new method is better?

Solution:

H₀: μ ≤ 70
H₁: μ > 70 (one-tailed)
Given: x̄ = 76, s = 8, n = 12, μ₀ = 70, α = 0.01
Test statistic: $t = \frac{x̄ - μ_{0}}{s / \sqrt{n}} = \frac{76 - 70}{8 / \sqrt{12}} = \frac{6}{8 / 3.464} = \frac{6}{2.31} = 2.60$
Degrees of freedom = n - 1 = 12 - 1 = 11
Critical value for α = 0.01 (one-tailed) with df = 11: $t_{α} = 2.718$
Since 2.60 < 2.718, we fail to reject H₀
Conclusion: There is insufficient evidence at the 1% significance level to conclude that the new teaching method is better.

Question 6:
Two groups of patients were given different treatments for the same condition. Group 1 consisted of 15 patients whose recovery times had a mean of 8.2 days with a standard deviation of 1.8 days. Group 2 consisted of 12 patients whose recovery times had a mean of 9.6 days with a standard deviation of 2.1 days. Test at the 5% significance level whether the mean recovery time for Group 1 is less than that for Group 2.

Solution:

H₀: μ₁ ≥ μ₂ (or μ₁ - μ₂ ≥ 0)
H₁: μ₁ < μ₂ (or μ₁ - μ₂ < 0)
Given: x̄₁ = 8.2, s₁ = 1.8, n₁ = 15, x̄₂ = 9.6, s₂ = 2.1, n₂ = 12, α = 0.05
Using pooled variance: $s_{p}^{2} = \frac{[(n_{1} - 1) s_{1}^{2} + (n_{2} - 1) s_{2}^{2}]}{n_{1} + n_{2} - 2} = \frac{[(15 - 1) (1.8)^{2} + (12 - 1) (2.1)^{2}]}{15 + 12 - 2} = \frac{[14 (3.24) + 11 (4.41)]}{25} = \frac{[45.36 + 48.51]}{25} = \frac{93.87}{25} = 3.75$
Test statistic: $t = \frac{{x̄}_{1} - {x̄}_{2}}{\sqrt{s_{p}^{2} (\frac{1}{n_{1}} + \frac{1}{n_{2}})}} = \frac{8.2 - 9.6}{\sqrt{3.75 (\frac{1}{15} + \frac{1}{12})}} = \frac{-1.4}{\sqrt{3.75 (0.0667 + 0.0833)}} = \frac{-1.4}{\sqrt{3.75 (0.15)}} = \frac{-1.4}{\sqrt{0.5625}} = \frac{-1.4}{0.75} = -1.87$
Degrees of freedom = n₁ + n₂ - 2 = 15 + 12 - 2 = 25
Critical value for α = 0.05 (one-tailed) with df = 25: $t_{α} \approx -1.708$
Since -1.87 < -1.708, we reject H₀
Conclusion: There is sufficient evidence to conclude that the mean recovery time for Group 1 is less than that for Group 2.

Question 7:
Two different machines are used to fill cereal boxes. A sample of 16 boxes from Machine A had a standard deviation of 15 grams. A sample of 12 boxes from Machine B had a standard deviation of 12 grams. At a 5% significance level, test whether the variance of the weights of boxes filled by Machine A is greater than that of Machine B.

Solution:

H₀: σ₁² ≤ σ₂² (or σ₁²/σ₂² ≤ 1)
H₁: σ₁² > σ₂² (or σ₁²/σ₂² > 1)
Given: s₁ = 15, n₁ = 16, s₂ = 12, n₂ = 12, α = 0.05
Test statistic: $F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{{(15)}^{2}}{{(12)}^{2}} = \frac{225}{144} = 1.5625$
Degrees of freedom: df₁ = n₁ - 1 = 16 - 1 = 15, df₂ = n₂ - 1 = 12 - 1 = 11
Critical value for α = 0.05 (one-tailed) with df₁ = 15, df₂ = 11: $F_{α} \approx 2.72$
Since 1.5625 < 2.72, we fail to reject H₀
Conclusion: There is insufficient evidence to conclude that the variance of weights of boxes filled by Machine A is greater than that of Machine B.

Question 8:
Two teaching methods are compared for their consistency in student performance. Method 1 was used with 21 students and resulted in a sample variance of 64. Method 2 was used with 25 students and resulted in a sample variance of 100. Test at the 1% significance level whether the two methods differ in their consistency (variance).

Solution:

H₀: $σ 1^{2} = σ 2^{2}$ (or $\frac{σ 1^{2}}{σ 2^{2}} = 1$ )
H₁: $σ 1^{2} \neq σ 2^{2}$ (or $\frac{σ 1^{2}}{σ 2^{2}} \neq 1$ )
Given: $s 1^{2} = 64$ , $n 1 = 21$ , $s 2^{2} = 100$ , $n 2 = 25$ , $α = 0.01$
Test statistic: $F = \frac{s 1^{2}}{s 2^{2}} = \frac{64}{100} = 0.64$
Degrees of freedom: $df 1 = n 1 - 1 = 20$ , $df 2 = n 2 - 1 = 24$
Critical value for α = 0.01 (two-tailed) with df₁ = 20, df₂ = 24: $F_{α / 2} \approx 2.66$
Critical value for α = 0.01 (two-tailed) with df₁ = 24, df₂ = 20: $F_{α / 2} \approx 0.375$
Since $0.64 > 0.375$ and $0.64 < 2.66$ , we fail to reject H₀
Conclusion: There is insufficient evidence to conclude that the two teaching methods differ in their consistency.

Question 9:
A die is rolled 120 times with the following frequencies: 1: 15, 2: 21, 3: 18, 4: 22, 5: 24, 6: 20. Test at a 5% level of significance whether the die is fair.

Solution:

H₀: The die is fair (equal probabilities of $\frac{1}{6}$ for each outcome)
H₁: The die is not fair
Given: Observed frequencies $O = [15, 21, 18, 22, 24, 20]$ , $n = 120$ , $α = 0.05$
Expected frequencies $E = n \times p = 120 \times \frac{1}{6} = 20$ for each outcome
Chi-square statistic: $χ^{2} = Σ \frac{{(O - E)}^{2}}{E} = \frac{{(15 - 20)}^{2}}{20} + \frac{{(21 - 20)}^{2}}{20} + \frac{{(18 - 20)}^{2}}{20} + \frac{{(22 - 20)}^{2}}{20} + \frac{{(24 - 20)}^{2}}{20} + \frac{{(20 - 20)}^{2}}{20} = \frac{25}{20} + \frac{1}{20} + \frac{4}{20} + \frac{4}{20} + \frac{16}{20} + 0 = 1.25 + 0.05 + 0.2 + 0.2 + 0.8 + 0 = 2.5$
Degrees of freedom $= k - 1 = 6 - 1 = 5$
Critical value for $α = 0.05$ with $df = 5$ : $χ_{α}^{2} = 11.07$
Since $2.5 < 11.07$ , we fail to reject H₀
Conclusion: There is insufficient evidence to conclude that the die is not fair.

Question 10:
A genetics theory predicts that the offspring of certain parents would have the following genotype distribution: 25% AA, 50% Aa, and 25% aa. In an experiment, 200 offspring were classified with the following results: 42 AA, 106 Aa, and 52 aa. At a 5% significance level, test whether the observed distribution follows the theoretical distribution.

Solution:

H₀: The distribution follows the theoretical probabilities $(0.25, 0.50, 0.25)$
H₁: The distribution does not follow the theoretical probabilities
Given: Observed frequencies $O = [42, 106, 52]$ , $n = 200$ , $α = 0.05$
Expected frequencies $E = n \times p = [200 \times 0.25, 200 \times 0.50, 200 \times 0.25] = [50, 100, 50]$
Chi-square statistic: $χ^{2} = Σ \frac{{(O - E)}^{2}}{E} = \frac{{(42 - 50)}^{2}}{50} + \frac{{(106 - 100)}^{2}}{100} + \frac{{(52 - 50)}^{2}}{50} = \frac{64}{50} + \frac{36}{100} + \frac{4}{50} = 1.28 + 0.36 + 0.08 = 1.72$
Degrees of freedom $= k - 1 = 3 - 1 = 2$
Critical value for $α = 0.05$ with $df = 2$ : $χ_{α}^{2} = 5.99$
Since $1.72 < 5.99$ , we fail to reject H₀
Conclusion: There is insufficient evidence to conclude that the observed distribution does not follow the theoretical distribution.

Question 11:
A market researcher believes that people's preference for four different brands of cereal follows the distribution: Brand A: 15%, Brand B: 35%, Brand C: 30%, Brand D: 20%. A survey of 300 consumers showed the following preferences: Brand A: 55, Brand B: 90, Brand C: 96, Brand D: 59. Test at a 1% significance level whether the observed preferences match the claimed distribution.

Solution:

H₀: The consumer preferences follow the claimed distribution $(0.15, 0.35, 0.30, 0.20)$
H₁: The consumer preferences do not follow the claimed distribution
Given: Observed frequencies $O = [55, 90, 96, 59]$ , $n = 300$ , $α = 0.01$
Expected frequencies $E = n \times p = [300 \times 0.15, 300 \times 0.35, 300 \times 0.30, 300 \times 0.20] = [45, 105, 90, 60]$
Chi-square statistic: $χ^{2} = Σ \frac{{(O - E)}^{2}}{E} = \frac{{(55 - 45)}^{2}}{45} + \frac{{(90 - 105)}^{2}}{105} + \frac{{(96 - 90)}^{2}}{90} + \frac{{(59 - 60)}^{2}}{60} = \frac{100}{45} + \frac{225}{105} + \frac{36}{90} + \frac{1}{60} = 2.22 + 2.14 + 0.4 + 0.017 = 4.78$
Degrees of freedom $= k - 1 = 4 - 1 = 3$
Critical value for $α = 0.01$ with $df = 3$ : $χ_{α}^{2} = 11.34$
Since $4.78 < 11.34$ , we fail to reject H₀
Conclusion: There is insufficient evidence to conclude that the observed preferences do not follow the claimed distribution.

Calculative Questions on Hypothesis Testing

Z-test for Single Mean

Z-test for Difference of Means

Student's t-test for Single Mean

Student's t-test for Difference of Means

F-test for Variances

Chi-Square Test for Goodness of Fit