AI HL · 13.4
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Hypothesis tests for a mean: mixed practice
13.4.14
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A decorator times how long two brands of paint take to dry, in minutes. Brand A is tested on 7 boards and brand B on 8 boards. Assume the drying times are normally distributed with equal variance. Test, at the 5% significance level, whether the mean drying times of the two brands are different.
Have a go first. Then open the working one line at a time.
Two independent samples, variance unknown and assumed equal: two-sample pooled t-test.
Different boards in each sample, so the samples are not paired. Use the pooled option.
H0: μ1 = μ2, H1: μ1 ≠ μ2, where μ1 and μ2 are the population mean drying times, in minutes, of brand A and brand B.
'Different' makes the test two-tailed.
Two-sample t-test, data lists, pooled, two-tailed: t = −4.19, p = 0.00105
Put each sample in its own list. The lists can be different lengths.
0.00105 < 0.05, so reject H0.
Compare p with the full 0.05, not with half of it.
There is sufficient evidence, at the 5% significance level, that the mean drying times of the two brands are different.
Say what it means for the paints. The sample mean of brand A is lower, so it dries faster in these samples.
The height gain of a variety of sunflower over four weeks is normally distributed with mean 12.4 cm and standard deviation 1.8 cm. A grower uses a new plant food and thinks the mean gain has increased. She assumes the standard deviation is unchanged and measures the gain of a random sample of 15 plants. (a) Write down suitable hypotheses. (b) Find the distribution of the sample mean \(\bar X\) when H0 is true. (c) Find the critical region at the 5% significance level. (d) The sample mean is 13.5 cm. State the conclusion.
Have a go first. Then open the working one line at a time.
H0: μ = 12.4, H1: μ > 12.4, where μ cm is the population mean height gain with the plant food.
She thinks the mean has increased, so the test is upper-tailed. σ is known, so this is a z-test.
Under H0, \(\bar X \sim N(12.4, 1.8^2/15)\), standard deviation \(1.8/\sqrt{15} = 0.465\)
The sample mean is less spread out than one plant: divide σ by √n.
From the calculator: inverse normal, area 0.95, mean 12.4, standard deviation 0.465: c = 13.2
5% of sample means lie above this value if H0 is true.
\(\bar X > 13.2\)
An upper region, because H1 has >.
13.5 > 13.2, so reject H0. There is sufficient evidence, at the 5% significance level, that the mean height gain has increased with the plant food.
The sample mean is in the critical region. The p-value, P(\(\bar X \geq 13.5\)) = 0.00897, agrees: it is below 0.05.
A jar of honey is labelled 454 g. The mass of a jar has mean 454 g and standard deviation 6.5 g when the filling machine works properly, but the distribution of masses is not known. An inspector weighs a random sample of 45 jars and tests, at the 5% significance level, whether the mean mass has changed. Assume the standard deviation is still 6.5 g. (a) Explain why the inspector can use a test based on the normal distribution. (b) Find the critical region. (c) The sample mean is 452.4 g. State the conclusion.
Have a go first. Then open the working one line at a time.
n = 45 is large, so by the central limit theorem \(\bar X\) is approximately normal, whatever the distribution of one jar.
The masses are not known to be normal, but the mean of a large sample is close to normal. A sample of 30 or more is usually enough.
H0: μ = 454, H1: μ ≠ 454
'Changed' is two-tailed: the region has two parts, with 2.5% in each tail.
Under H0, \(\bar X \approx N(454, 6.5^2/45)\), standard deviation = 0.969
Divide σ by √n.
From the calculator: inverse normal, areas 0.025 and 0.975, mean 454, standard deviation 0.969: 452.1 and 455.9
One value below 454 and one above it.
\(\bar X < 452.1\) or \(\bar X > 455.9\)
Reject H0 if the sample mean lies in either part.
452.4 is between 452.1 and 455.9, so do not reject H0. There is insufficient evidence, at the 5% significance level, that the mean mass of a jar has changed.
452.4 is not in the critical region.
The yield of wheat, in tonnes per hectare, is normally distributed. For the old variety Y the standard deviation is 0.4 and for the new variety X it is 0.5, and both are known. A farmer believes that variety X gives a mean yield more than 0.6 tonnes per hectare greater than variety Y. She grows X on 30 fields, with mean yield 6.92, and Y on 40 fields, with mean yield 6.15. (a) Explain why a two-sample z-test can be used. (b) Test her belief at the 5% significance level.
Have a go first. Then open the working one line at a time.
The two samples are independent, the populations are normal and both standard deviations are known.
Known σ means a z-test, not a t-test. The samples are different fields, so they are not paired.
H0: μX − μY = 0.6, H1: μX − μY > 0.6, where μX and μY are the population mean yields of X and Y.
H0 states the difference she claims to beat. The hypothesised difference is not zero.
Under H0, \(\bar X_X - \bar X_Y \sim N\left(0.6,\ \dfrac{0.5^2}{30} + \dfrac{0.4^2}{40}\right)\), standard deviation = 0.111
The variances of independent sample means add, even for a difference.
Observed difference = 6.92 − 6.15 = 0.77. From the calculator: normal cdf from 0.77 to ∞, mean 0.6, standard deviation 0.111: p = 0.0629
Upper tail, because H1 has >.
0.0629 > 0.05, so do not reject H0. There is insufficient evidence, at the 5% significance level, that variety X gives a mean yield more than 0.6 tonnes per hectare greater than variety Y.
The sample difference, 0.77, is bigger than 0.6, but not by enough to rule out chance.
A coach thinks that running on grass is slower than running on the track. Nine athletes each run 100 m on the track (T) and on grass (G). The times, in seconds, are shown. Each time is normally distributed with standard deviation 0.3 s, and the times are independent. Let D = G − T. (a) Write down hypotheses and form the differences D. (b) Find the variance of D, and the distribution of \(\bar D\) when H0 is true. (c) Find the critical region at the 5% significance level and state the conclusion.
Have a go first. Then open the working one line at a time.
D = G − T: 0.3, 0.1, 0.4, 0.2, 0.3, 0.1, 0.4, 0.3, 0.1. H0: μD = 0, H1: μD > 0, where μD s is the population mean of G − T.
The coach thinks grass is slower, so G − T is positive on average. Test the differences, not the two samples.
Var(D) = Var(G) + Var(T) = 0.3² + 0.3² = 0.180
Var(G − T) = Var(G) + Var(T) for independent times. The variances add, they are not subtracted.
Under H0, \(\bar D \sim N(0, 0.180/9)\), standard deviation = 0.141
Divide the variance of D by n = 9.
From the calculator: inverse normal, area 0.95, mean 0, standard deviation 0.141: c = 0.233. Critical region: \(\bar D > 0.233\)
Upper tail, because H1 has >.
\(\bar D = 0.244\) > 0.233, so reject H0. There is sufficient evidence, at the 5% significance level, that the mean time on grass is greater than on the track.
The sample mean difference is in the critical region. The p-value is 0.0420, which is below 0.05.
A typing course claims to raise speed. The typing speeds, in words per minute, of 8 students are recorded before and after the course. (a) Explain why a paired test is suitable. (b) Form the differences, after − before, and find their mean and standard deviation. (c) Test, at the 5% significance level, whether the course raises the mean speed. State the assumption you make.
Have a go first. Then open the working one line at a time.
The same 8 students are measured twice, so each 'before' belongs with one 'after'. The two samples are not independent.
Pairing removes the large differences in speed between students and leaves the effect of the course.
D = after − before: 3, 1, 2, 0, 4, 1, 3, 1. Mean = 1.88, s = 1.36
From the calculator, one-variable statistics of the list D. s is the sample standard deviation.
H0: μD = 0, H1: μD > 0, where μD is the population mean increase in speed, in words per minute.
The course should raise the speed, so after − before is positive.
t-test on the list D (or paired t-test), μ0 = 0, upper tail: t = 3.91, p = 0.00291
The calculator gives t and p for the upper tail.
0.00291 < 0.05, so reject H0. There is sufficient evidence, at the 5% significance level, that the course raises the mean typing speed.
Compare p with 0.05.
The population of differences is normally distributed.
For a pooled two-sample test you would assume two normal populations with equal variance. A paired test needs only one assumption: the differences are normal.
Mia believes that her new cycling route to school is more than 4 minutes quicker, on average, than the old route. She times 10 journeys on the old route (mean 34.8 minutes, sample variance 9.6) and 12 journeys on the new route (mean 29.1 minutes, sample variance 7.2). Assume the journey times are normally distributed with equal variance. Test her belief at the 10% significance level.
Have a go first. Then open the working one line at a time.
s1² = 10/9 × 9.6 = 10.7, s2² = 12/11 × 7.2 = 7.85
A sample variance divides by n. The estimate of the population variance multiplies it by n/(n − 1).
H0: μ1 − μ2 = 4, H1: μ1 − μ2 > 4, where μ1 and μ2 minutes are the population mean journey times on the old and new route.
Old route is group 1. 'More than 4 minutes quicker' means the old mean is more than 4 greater.
sp² = (9 × 10.7 + 11 × 7.85) / 20 = 9.12
Combine the two estimates, weighting each by its degrees of freedom. Total degrees of freedom = 10 + 12 − 2 = 20.
t = (34.8 − 29.1 − 4) / √(9.12 × (1/10 + 1/12)) = 1.31
The calculator's two-sample test only tests a difference of 0, so use it for the pooled estimate and form t yourself.
From the calculator: t cdf, 20 degrees of freedom, upper tail from 1.31: p = 0.102
Upper tail, because H1 has >.
0.102 > 0.1, so do not reject H0. There is insufficient evidence, at the 10% significance level, that the new route is more than 4 minutes quicker, on average.
Compare p with 0.10.