Question

A consumer products testing group is evaluating two competing brands of tires, Brand 1 and Brand 2. Tread wear can vary considerably depending on the type of car, and the group is trying to eliminate this effect by installing the two brands on the same random sample of 8 cars. In particular, each car has one tire of each brand on its front wheels, with half of the cars chosen at random to have Brand 1 on the left front wheel, and the rest to have Brand 2 there. After all of the cars are driven over the standard test course for 20,000 miles, the amount of tread wear (in inches) is recorded, as shown in Table 1. Difference Car Brand 1 Brand 2 (Brand 1. Brand 2) 1 0.098 0.042 0 .262 0.291 0.220 0.399 0.283 0.269 0.369 0.336 0.164 0.249 0.144 0.356 0.263 0.115 0.154 0.191 0.043 0.020 5 6 0.154 0.215 8 0.145 Table 1 Based on these data, can the consumer group conclude, at the 0.05 level of significance, that the mean tread wears of the brands differ? Answer this question by performing a hypothesis test regarding (which is u with a letter "d" subscript), the population mean difference in tread wear for the two brands of tires. Assume that this population of differences (Brand 1 minus Brand 2) is normally distributed. Perform a two-talled test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.) The null hypothesis: HOP

Answer 1

The hypothesis can be constructed as:

Null hypothesis, ${\mu _d} = 0$ ( There is no significance difference between mean tread wears of Brand 1 and Brand 2.)

Alternative hypothesis, ${\mu _d} \ne 0$ (There is a significance difference between mean tread wears of Brand 1 and Brand 2.)

The type of test statistic is two tailed.

From the provided information sample size (n) = 8

The degree of freedom = n -1

df = 8 - 1

df = 7

The mean of difference can be calculated as:

$\begin{array}{c}\bar d = \frac{{\sum\limits_{i = 1}^n {{d_i}} }}{n}\\ = \frac{{0.098 + \cdots + 0.145}}{8}\\ = 0.099\end{array}$

The standard deviation of difference can be calculated as:

$\begin{array}{c}{s_d} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({d_i} - \bar d)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {0.098 - 0.099} \right)}^2} + \cdots + {{\left( {0.145 - 0.099} \right)}^2}}}{{8 - 1}}} \\ = 0.067\end{array}$

The paired t test will use to calculate the test statistic.

Now, the test statistic can be calculated as:

$\begin{array}{c}t = \frac{{\bar d}}{{\frac{{{s_d}}}{{\sqrt n }}}}\\ = \frac{{0.099}}{{\frac{{0.067}}{{\sqrt 8 }}}}\\ = 4.179\end{array}$

The critical value at 0.05 level of significance with 7 degree of freedom from t value table is 2.365.

The test statistic 4.179 is greater than the critical value 2.365, therefore the null hypothesis is rejected and it can be concluded that there is a significance difference between mean tread wear of Brand 1 and Brand 2.