Lavender No Lavender Minutes Euro Spent Minutes Euro Spent 103 149 92 22.50 68 1

Question

Lavender No Lavender Minutes Euro Spent Minutes Euro Spent 103 149 92 22.50 68 174 126 19.00 79 149 114 22.30 174 106 116 22.40 20.30 21.9 137 121 24.00 92 149 933 20.50 74 149 76 23.00 172 149 97 22.43 72 149 107 21.90 74 149 124 21.40 59 174 104 19.00 63 174 129 24.40 77 174 103 20.30 109 20.4 117 21.00 174 109 21.90 104 18.00 104 20.30 149 112 66 24.00 86 149 107 21.90 174 107 94 24.00 21.90 18.4. 121 82 149 109 20.00 97 18.4. 114 19.40 93 149 116 21.00 108 87 149 109 21.90 101 24.4 97 20.70 75 129 111 21.90 87 149 109 21.90 101 24.4 97 21.00 86 14.9 106 21.40

Explanation / Answer

Part a

Here, we have to use two sample t test for the population mean. We have to check the claim whether the use lavender odour encourage the customers to stay longer in the restaurant or not. Here, we have to use two sample t test because we are not given the population standard deviations for the given two populations. The null and alternative hypothesis for this test is given as below:

H0: µ1 = µ2 versus Ha: µ1 < µ2

We consider = 0.05

From the given data, we have

X1bar = 86.375

X2bar = 106.5938

S1 = 15.76388

S2 = 13.72744

N1 = 32

N2 = 32

Degrees of freedom = 32 + 32 – 2 = 62

Test statistic formula is given as below:

t = (X1bar –X2bar)/sqrt[(S1^2/N1)+(S2^2/N2)]

t = (86.375 – 106.5938)/sqrt[(15.76388^2/32)+(13.72744^2/32)]

t = -5.4716

Critical value = -1.6698

P-value = 0.00

= 0.05

P-value < = 0.05

So, we reject the null hypothesis

This means we conclude that there is sufficient evidence that the use lavender odour encourage the customers to stay longer in the restaurant.

Part b

Now, we have to check whether the use of Lavender odour increases the total spend of customer or not. We have to use the two sample t test for the population mean because we are not given the population standard deviations. We consider 5% level of significance for this test. The null and alternative hypotheses are given as below:

H0: µ1 = µ2 versus Ha: µ1 < µ2

We consider = 0.05

From the given data, we have

X1bar = 16.82188

X2bar = 21.40719

S1 = 2.738935

S2 = 1.566991

N1 = 32

N2 = 32

Degrees of freedom = 32 + 32 – 2 = 62

Test statistic formula is given as below:

t = (X1bar –X2bar)/sqrt[(S1^2/N1)+(S2^2/N2)]

t = (16.82188 - 21.40719)/sqrt[(2.738935^2/32)+( 1.566991^2/32)]

t = -8.2201

Critical value = -1.6698

P-value = 0.00

= 0.05

P-value < = 0.05

So, we reject the null hypothesis

This means we conclude that there is sufficient evidence that the use lavender odour encourage the customers to spend more in the restaurant.

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