1501:25:21 A manufacturer claims that the calling range (in feet) of its 900-MHz

Question

1501:25:21 A manufacturer claims that the calling range (in feet) of its 900-MHz cordless telephone is greater than that of its leading competitor. A sample of 9 phones from the manufacturer had a mean range of 1100 feet with a standard deviation of 44 feet. A sample of 18 similar phones from its competitor had a mean range of 1040 feet with a standard deviation of 25 feet. Do the results support the manufacturer's claim? Let be the true mean range of the manufacturers cordless telephone and 2 be the true mean range of the competitors cordless telephone. Use a significance level of = 0.1 for the test. Assume that the population variances are equal and that the two populations are normally distributed. Step 1 of 4: State the null and alternative hypotheses for the test. Keypad Tables Answer 10 Points 18 Hawkes Learning

Explanation / Answer

xbar1=1100

xbar2=1040

s1=44

s2=25

n1=9

n2=18

Degrees of freedom = n1 + n2 - 2 = 25

Level of significance = 0.1

1)

Ho: 1 = 2

Ha: 1 > 2

2)

the t-statistic is computed as follows:

t= ( xbar1 - xbar2 ) ÷ { [( (n1 - 1)*s1² + (n2 - 1)*s2²) / (n1 + n2 - 2)] × [ (1/n1) + (1/n2) ] }

t = ( 1100 - 1040 ) ÷ { [( ( 9 - 1)*44² + ( 18-1)*25²) / ( 25 ) ] × [ (1/9) + (1/18)] }

t = 4.547

3)

reject Ho if

t statistic > t critical

t=4.547 > tc=1.316, ( tc is t critical value at 0.1 level of significance for degrees of freedom 25)

the null hypothesis is rejected.

Using the P-value approach:

The p-value p=0.0001<0.1,

null hypothesis is rejected

4)

We reject the null hypothesis.

We can concluded that there is enough evidence to claim that population mean 1 is greater than 2, at the 0.1 significance level.

Cordless telephone is greater than its leading competitor.

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