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.

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

Step-by-step 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 = average amount spent on gifts celebrating Valentine's Day

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

Alternate Hypothesis, : > $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.  =  ~

where, = sample average amount spent = $148.50

             s = sample standard deviation = $34.90

             n = sample of people = 18

So, test statistics  =    ~

                                    =  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( > 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.

answer: .64

step-by-step explanation

set up the formula for standard deviation for this set of numbers, then simplify