Comparing Two Means Hypothesis Testing

Describe the kind of hypothesis test would you use in the following scenarios.

1. A mobile phone company has released a new software update. They have asked you to find statistical evidence to support their claim that the software update has improved battery life.

2. A pet store sells Welsh Corgi puppies from two different breeders. They wish to determine whether there is a significant difference between the weights of the puppies from each breeder.

Solution

1. In order to conduct this experiment, you would need to collect samples of information on phone battery life before and after the software update. Since the samples will be taken from the same population after a change has been made, they are not independent. Therefore, you need to use a paired t-test.

2. In this case, you would be required to take samples of weights from two different breeders and therefore two independent distributions. You should assume that the populations have the same variances, therefore you will need to use a pooled estimate of variance to find the t-value and not a paired t-test.

A pet store sells Welsh Corgi puppies on behalf of two puppy breeders, \(X\) and \(Y\). They have sampled the weights of puppies from each breeder.

Fig. 1 - Puppies always make math better!

Weights of puppies from breeder \(X\) in kilograms: \(5.44,5.32,5.21,5.67.\)

Weights of puppies from breeder \(Y\) in kilograms: \(5.02,4.99,5.42,5.21,5.11.\)

The pet store wishes to know whether there is a statistically significant difference between the weights of the puppies from each breeder.

a. If you wanted to test the difference in the weights of the puppies, what assumptions need to be made?

b. Test whether the mean weights of puppies from the two breeders is different at the \(10\%\) confidence level.

Solution

a. In order to test the difference in the weights of the puppies, the assumptions to be made are that the samples of puppies are normally distributed, independent and have the same variances.

b. The test is two-tailed, so the hypotheses are,

\[ \begin{align} &H_0:\, \mu _x=\mu _y \\ &H_1: \,\mu _x \neq \mu _y.\end{align}\]

This is a two-tailed test since the alternative hypothesis is that the mean weights are different. The significance level is \(10\)%, so the critical region will have the probability of \(0.05\) in each tail of the distribution.

The number of degrees of freedom is

\[\upsilon = (4-1)+(5-1)=7.\]

The critical value can be found using a calculator or probability tables:

\[t_{\upsilon =7}(0.05)=1.895.\]

Next, find the pooled estimate of variance. You should have \(\bar{x}=5.41\) and \(\bar{y}=5.17.\)

The samples variances are \(s^2_x=0.038866667 \) and \(s^2_y=0.03015\).

Therefore, the pooled estimate of variance is,

\[\begin{align} s^2_p &= \frac{(n_x-1)s^2_x+(n_y-1)s^2_y}{(n_x-1)+(n_y-1)} \\&= \frac{(4-1)0.038867 +(5-1)0.03015 }{(4-1)+(5-1)} \\&=0.033886 \text{ to 5 s.f.} \end{align}\]

Your value of \(t^*\) is then:

\[\begin{align} t&=\frac{(\bar{x}-\bar{y})-(\mu _x - \mu _y)}{\sqrt{s^2_p\left(\dfrac{1}{n_x}+\dfrac{1}{n_y}\right)}}\\&=\dfrac{(5.41-5.17)-(0)}{\sqrt{0.033886\left(\dfrac{1}{4}+\dfrac{1}{5}\right)}}\\&=1.9435\end{align}\]

Since \(t^*=1.9435>1.895=t_\upsilon\), your value of \(t^*\) falls within the critical region. Therefore, at the \(10\)% significance level, you can reject the null hypothesis.

In conclusion, there is evidence to suggest there is a difference between the means of the weights of Welsh Corgi puppies from the two breeders.

A food delivery service, \(A\), claims that their average food delivery time is more than \(5\) minutes faster than the delivery time of their competitor, \(B\).

A random sample of delivery times from each company is collected:

• Food delivery time for \(A\), in minutes: \(22,16,45,23,39,32.\)
• Food delivery time for \(B\), in minutes: \(34,42,63,18,25,46,47.\)

Food delivery service \(B\) hires you to test whether this claim is statistically significant at the \(10\%\) significance level. Complete a hypothesis test for the difference between means and explain what this means for the two food delivery services.

Solution

Since the samples are independent the null hypothesis would normally be that the two means are the same. However the claim is that service \(A\) averages \(5\) minutes faster than their competitor, so the null hypothesis is instead \(\mu _A=\mu _B -5 \). Since you are only interested in whether the food delivery time is greater for one service, the hypotheses are:

\[ \begin{align} &H_0:\,\mu _A=\mu _B -5 \\ &H_1: \,\mu_A < \mu _B-5. \end{align}\]

This is a one-tailed test.The significance level is \(10\)%, so the critical region will have the probability of \(0.10\) in the left tail of the distribution.

The number of degrees of freedom are

\[\upsilon = (6-1)+(7-1)=11.\]

The critical value can be found using a calculator or probability tables,

\[t_{\upsilon =11}(0.10)=1.363.\]

Since you are only interested in whether \(\mu _a\) is less than \(\mu _b -5\), the critical value is \(t_\upsilon = -1.363\).

Next, find the pooled estimate of variance. You have \(\bar{a}=29.5\) and \(\bar{b}=39.3\). The samples variances are \(s^2_a=123.50 \) and \(s^2_b=226.57\). Therefore, the pooled estimate of variance is:

\[\begin{align} s^2_p &= \frac{(n_a-1)s^2_a+(n_b-1)s^2_b}{(n_a-1)+(n_b-1)} \\&= \frac{(6-1)123.50 +(7-1)226.57 }{(6-1)+(7-1)} \\&=179.72\text{ to 5 s.f.} \end{align}\]

The value of \(t^*\) is therefore,

\[\begin{align} t^*&=\frac{(\bar{a}-\bar{b})-(\mu _a - \mu _b)}{\sqrt{s^2_p\left(\dfrac{1}{n_a}+\dfrac{1}{n_b}\right)}}\\&=\dfrac{(29.5 -39.3)-(-5)}{\sqrt{179.72 \left(\dfrac{1}{6}+\dfrac{1}{7}\right)}}\\&=-0.64357.\end{align}\]

Since \(t^*=-0.64357>-1.363=t_\upsilon \), the value of \(t\) falls within the acceptance region. Therefore, at the \(10\%\) significance level, you fail to reject the null hypothesis.

This means that there is not sufficient evidence to suggest that delivery service \(A\) has a delivery time better than \(5\) minutes faster than delivery service \(B\).