Question

Hallmark would like to test the hypothesis that those celebrating Valentine's Day will spend more than an average of $125 on gifts. A random sample of 18 people celebrating Valentine's Day spent an average of $148.50 with a standard deviation of $34.90. Hallmark would like to set α = 0.01. Use the p-value approach to test this hypothesis.

Answer 1

We conclude that those celebrating Valentine's Day spend more than an average of $125 on gifts.

Explanation:

We are given that Hallmark would like to test the hypothesis that those celebrating Valentine's Day will spend more than an average of $125 on gifts.

A random sample of 18 people celebrating Valentine's Day spent an average of $148.50 with a standard deviation of $34.90.

Let
`\mu` = average amount spent on gifts celebrating Valentine's Day

So, Null Hypothesis,
`H_0` :
`\mu`
`\leq` $125 {means that those celebrating Valentine's Day spend less than or equal to an average of $125 on gifts}

Alternate Hypothesis,
`H_A` :
`\mu` > $125 {means that those celebrating Valentine's Day spend more than an average of $125 on gifts}

The test statistics that will be used here is One-sample t test statistics as we don't know about population standard deviation;

T.S. =
`(\bar X-\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`\bar X` = sample average amount spent = $148.50

s = sample standard deviation = $34.90

n = sample of people = 18

So, test statistics =
`(148.50-125)/((34.90)/(√(18) ) )` ~
`t_1_7`

= 2.857

The value of the sample test statistics is 2.857.

Now, P-value of the test statistics is given by the following formula;

P-value = P(
`t_1_7` > 2.857) = 0.0056

Because in the t table the critical value of 2.857 at 17 degree of freedom will lie between P = 1% and P = 0.5%.

Now, since P-value of test statistics is less than the level of significance as 0.01 > 0.0056, so we sufficient evidence to reject our null hypothesis.

Therefore, we conclude that those celebrating Valentine's Day spend more than an average of $125 on gifts.