Question

A restaurant recently began offering free dessert on Monday nights with the order of any dinner menu item. Before the offer, the mean number of dinner customers on Monday nights was 106. Following are the numbers of diners on 12 different days after the offer went into effect.

98	121	124	112	94	108
128	118	123	120	119	124

Using a significance level of α = 0.05, can we conclude that the mean number of diners on Monday night increased while the offer of free dessert was in effect?

a.) State the null and alternative hypothesis using correct symbolic form.

H₀: Answerρμσ Answer=≠<> Answer

H₁: Answerρμ σ Answer≠<> Answer

b.) Is this a left-tailed, right-tailed, or two-tailed hypothesis test?

left-tailed

right-tailed

two-tailed

c.) How many degrees of freedom will be used for the given sample size?

Answer

d.) What is the positive critical value? (round to two decimal places)

t = Answer

e.) What is the test statistic? (round to two decimal places)

t = Answer

f.) What is the p-value? (round to four decimal places)

p-value is Answer

g.) Should we reject or fail to reject the null hypothesis?

reject

fail to reject

h.) State the conclusion.

There is sufficient evidence to support the claim that the mean number of diners on Monday night increased while the offer of free dessert was in effect.

There is not sufficient evidence to support the claim that the mean number of diners on Monday night increased while the offer of free dessert was in effect.

There is sufficient evidence to warrant rejection that the mean number of diners on Monday night increased while the offer of free dessert was in effect.

There is not sufficient evidence to warrant rejection that the mean number of diners on Monday night increased while the offer of free dessert was in effect.

Answer 1

Let be the true mean number of diners on Monday night while the offer of free dessert was in effect. Before the offer, the mean number of dinner customers on Monday nights was 106. We want to test if the mean number of diners on Monday night increased while the offer of free dessert was in effect. That is, we want to test if . This is the alternative hypothesis as alternative hypothesis always has one of these inequality:

We have the following information from the sample

n=12 is the sample size

The sample mean number of customers on Monday night during the offer period is

the sample standard deviation is

We will estimate the population standard deviation using the sample.

The estimated population standard deviation is

The estimated standard error of mean is

a.) State the null and alternative hypothesis using correct symbolic form.

ans:

H0: μ = 106

H1: μ > 106

b.) Is this a left-tailed, right-tailed, or two-tailed hypothesis test?

The alternative hypothesis has ">", hence this is a right tailed test

ans: right-tailed

c.) How many degrees of freedom will be used for the given sample size?

The sample size n=12. The degrees of freedom are n-1=12-1=11

Answer: 11

d.) What is the positive critical value? (round to two decimal places)

The sample size, n=12 is less than 30 and we do not know the population standard deviation. Assuming a normal distribution for number of customers on Monday nights, we can say that the sampling distribution of mean is t distribution. That is we will use 1 sample t test for means

The significance level is α = 0.05.

The right tail critical value is (This is a right tailed test and hence the area under the right tail=α = 0.05.

Using the t tables for df=11, and the area under the right tail = 0.05, we get

ans: tcrit = 1.80

e.) What is the test statistic? (round to two decimal places)

The hypothesized value of mean is

The test statistic is

ans: t = 3.15

f.) What is the p-value? (round to four decimal places)

this is a right tailed test. The p-value is the area under the right tail to the right of test statistic

The degrees of freedom is n-1=12-1=11

We need to use technology (calculator or Excel function, =T.DIST.RT(3.153,11)), to get p-value=0.0046

ans: p-value is 0.0046

g.) Should we reject or fail to reject the null hypothesis?

Using the critical value method:

We will reject the null hypothesis if the test statistic is greater than the critical value.

Here, the test statistic is 3.15 and it is greater than the critical value 1.80. Hence, we reject the null hypothesis.

Using the p-value method:

We will reject the null hypothesis if the p-value is less than the significance level alpha=0.05.

Here, the p-value is 0.0046 and it is less than the significance level alpha=0.05. Hence, we reject the null hypothesis.

ans: reject

h.) State the conclusion.

ans:

There is sufficient evidence to support the claim that the mean number of diners on Monday night increased while the offer of free dessert was in effect.