Hi - could someone please assist me with this question? It\'s giving me nightmar

Question

Hi - could someone please assist me with this question? It's giving me nightmares......thanks

Historically, the average waiting time spent in a queue on Friday afternoon at a bank's city branch was 10 minutes. To determine whether their new multi-queuing system constitutes an improvement, a random sample of 15 customers was taken and their waiting time recorded. The results are in the X1 column of the datafile P14.12. Assume the distribution of waiting times is normally distributed.

1. State the direction of the alternative hypothesis used to test whether there is an improvement in average waiting time. Type gt (greater than), ge (greater than or equal to), lt (less than), le (less than or equal to) or ne (not equal to) as required in the box.

2. Calculate the test statistic correct to three decimal places (hint: use Descriptive Statistics to calculated the standard deviation and sample mean).

3. Use Kaddstat to determine the p-value for the test correct to four decimal places.

4. Is the null hypothesis rejected for this test if a 10% level of significance is used? Type yes or no.

5. Disregarding your answer for 4, if the null hypothesis was rejected, could we conclude that the average waiting time has improved with the new queuing system? Type yes or no.

DATA FILE

Y            X1                  X2 28 12.6 134 43 11.4 126 45 11.5 143 49 11.1 152 57 10.4 143 68 9.6 147 74 9.8 128 81 8.4 119 82 8.8 130 86 8.9 135 101 8.1 141 112 7.6 123 114 7.8 121 119 7.4 129 124 6.4 135

Explanation / Answer

1. From the data you have given here, we find that the average values from column X1 is 9.32. Since you want to test whether the new systems causes an improvement, you have to propose the alternate hypothesis as

H1: ?<10

Type lt (as per your question).

2. Test stastic calculation involves finding out the 't' value since the number of samples is less and the population variance is unknown We need to use the sample variance/sample standard deviation.

From the data given, Sample standard deviation= 1.78774

t= x- ?/(s/sqrtn)

Sample mean x=9.32,

Population mean ?=10

Sample standard deviation= 1.78774

Number of samples=n=15

Test statistic t=9.32-16/(1.78774/sqrt(15))

t=-14.4716

3. p=0.163

4. At 10% level of significance, we find the value of t at 90% level confidence,

Degrees of freedom=15-1=14

t at 0.9, 14 degrees of freedom=1.345

Since the t distribution is normal and symmetric, t value is either 14.4716/-14.4716

Since p value > 0.1, we accept null hypothesis and reject alternate hypothesis at 90% level of confidence.

5. No, the average waiting time has not reduced with the new system. This is because we accept our

alternate hypothesis since we do not have sufficient evidence to accept null hypothesis, that doesn't mean that the new system has improved waiting time.

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