The following table shows the results of two random samples that measured the av

Question

The following table shows the results of two random samples that measured the average number of minutes per charge for AA? Lithium-ion (Li-ion) rechargeable batteries versus?Nickel-Metal Hydride? (NiMH) rechargeable batteries. Complete parts a through c.

NiMH 82.8 11.4 19 Li-ion 96.7 Sample mean Sample standard deviation Sample size 13 a. Perform a hypothesis test using = 0.05 to determine if the average number of minutes per charge differs between these two battery types. Assume the population variances for the number of minutes per charge are not equal Determine the null and alternative hypotheses for the test. Calculate the appropriate test statistic and interpret the result. The test statistic is (Round to two decimal places as needed.) The critical value(s) is(are) (Round to two decimal places as needed. Use a comma to separate answers as needed.) Because the test statistic the null hypothesis

Explanation / Answer

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   =   0  
Ha:   u1 - u2   =/   0   [ANSWER]

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At level of significance =    0.05          

As we can see, this is a    two   tailed test.      
Calculating the means of each group,              
              
X1 =    96.7          
X2 =    82.8          
              
Calculating the standard deviations of each group,              
              
s1 =    6.1          
s2 =    11.4          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    13          
n2 = sample size of group 2 =    19          
Thus, df = n1 + n2 - 2 =    30          
Also, sD =    3.114852756          
              
Thus, the t statistic will be              
              
t = [X1 - X2 - uD]/sD =    4.462490233   [ANSWER, TEST STATISTIC]

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where uD = hypothesized difference =    0          
              
Now, the critical value for t is, as this is two tailed,              
              
tcrit = -2.042272456, 2.042272456   [ANSWER]

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**Because the test statistic FALLS WITHIN THE CRITICAL VALUES - REJECTS the null hypothesis. [ANSWR]

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b)
               
Also, using p values, as this is two tailed.              
              
p =    0.000105738   [ANSWER, P VALUE]

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OPTION A. Since the P value is less than the significance level, we reject Ho.

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OPTION C: It must be assumed that the distributions for both populations are normally distributed, and the samples are independent.

      

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Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!

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