Question

A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a less than desirable battery life. They aim to compete with a new touch phone that is guaranteed to have a battery life more than two hours longer than the leading product. A recent sample of 120 units of the leading product provides a mean battery life of 5 hours and 40 minutes with a standard deviation of 30 minutes. A similar analysis of 100 units of the new product results in a mean battery life of 8 hours and 5 minutes and a standard deviation of 55 minutes. It is not reasonable to assume that the population variances of the two products are equal. Let new products and leading products represent population 1 and population 2, respectively.
Sample 1 is from the population of new phones and Sample 2 is from the population of old phones. All times are converted into minutes.

a. Set up the hypotheses to test if the new product has a battery life more than two hours longer than the leading product.
b-1. Calculate the value of the test statistic. (Round intermediate calculations to 4 decimal places and final answer to 2 decimal places.)
b-2. Implement the test at the 5% significance level using the critical value approach.

Answer 1

Answer:

a) Null hypothesis:$\mu_{1} - \mu_{2} \leq 120$

Alternative hypothesis:$\mu_{1} - \mu_{2}>120$

b) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{(485-340)-120}{\sqrt{\frac{55^2}{100}+\frac{30^2}{120}}}}=4.0689$  

c) Critical value

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=100+120-2=218$

The significance level on this case is 0.05 or 5% so we need to find a quantile on the t distribution with 218 degrees of freedom that accumulates 0.95 of the area on the left and 0.05 of the area on the right tail and this value can be founded with the following excel code: "=T.INV(1-0.05,218)" and we got:

$t_{crit}= 1.652$

Conclusion

Since our calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance and then we can conclude that the new product has a battery life more than two hours (120 minutes) longer than the leading product

Step-by-step explanation:

Data given and notation

$\bar X_{1}= 8*60 +5 =485$ represent the mean for the new sample

$\bar X_{2}=5*60+40=340$ represent the mean for the old sample

$s_{1}=55$ represent the sample standard deviation for the new sample

$s_{2}=30$ represent the sample standard deviation for the old sample

$n_{1}=100$ sample size selected for the new sample

$n_{2}=120$ sample size selected for the old sample

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

a) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the new product has a battery life more than two hours longer than the leading product., the system of hypothesis would be:

Null hypothesis:$\mu_{1} - \mu_{2} \leq 120$

Alternative hypothesis:$\mu_{1} - \mu_{2}>120$

If we analyze the size for the samples both are higher than 30 but we don't know the population deviations so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})- \Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

b) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{(485-340)-120}{\sqrt{\frac{55^2}{100}+\frac{30^2}{120}}}}=4.0689$  

c) Critical value

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=100+120-2=218$

The significance level on this case is 0.05 or 5% so we need to find a quantile on the t distribution with 218 degrees of freedom that accumulates 0.95 of the area on the left and 0.05 of the area on the right tail and this value can be founded with the following excel code: "=T.INV(1-0.05,218)" and we got:

$t_{crit}= 1.652$

Conclusion

Since our calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance and then we can conclude that the new product has a battery life more than two hours (120 minutes) longer than the leading product