Question

The inter-arrival times of customers arriving at a HDB coffee shop were found to follow Exponential distribution L H Frequency The grouped data shows frequency counts of sampled inter- 0 1 41 arrival times occurring greater than or equal to L minutes and 1 2 21 less than H minutes. A total of 100 sample inter-arrival times 23 19 were measured. Use the mid-point as a representative data in calculations for each interval. Test the claim that the population mean inter-arrival time is equal to 2 minutes at 2.5% significance level. Calculate the 8 9 1 p-value of the test statistic, and hence use p-value to 9 10 conclude the hypothesis test.

Answer 1

Here we enter these data into Excel with taking mid-point for each interval. And take mid point as a representative data in calculations for each interval.

Here we test that, the claim that the population mean inter-arrival time is equal to 2 minutes at 2.5% significance level.

Here Hypothesis is as

Null Hypothesis : H_o = mean inter-arrival time is equal to 2 min.

V/S Alternative Hypothesis : H₁ = mean inter-arrival time is not equal to 2 min.

Test Statistic :

here sample size is 9 (<30) so we use t-test as

$t = \frac{\bar{x}-\mu}{S/\sqrt{N}}$

Let,

$\bar{x}= \frac{\sum X*F}{N} = \frac{194}{100} = 1.94$

$S^{2}= \frac{\sum X^{2}*F - \frac{(\sum X*F)^{2}}{N}}{N-1} = \frac{683 - 376.36}{99} = 3.0974$

S = 1.76

N = 100 and $\mu = 2$

Let, $t = \frac{1.94 - 2}{1.76/10} = -0.34$

Calculate p value as using EXCEL formula

p = TDIST(ABS(t) , N-1 , 2) = TDIST(0.34 , 99 , 2) = 0.7346

p = 0.7346

Here given significance level is 0.025. here p_calculated > 0.025

So we accept null hypothesis, That is population mean inter-arrival time is equal to 2 minutes.