In a chocolate chip cookie investigation, you decide to compare two cookie sources. Source 1: Mom mails to you a package containing a dozen of her homemade chocolate chip cookies. Source 2: You purchase a dozen of a store brand of chocolate chip cookies at the convenience store near campus. (Assume that the two types of cookies are the same size, same weight, etc.) You believe that Mom’s cookies have more chips. You eat the cookies, carefully counting the number of chips with the following results: Mom’s cookies: n = 12, mean = 9, s = 3; Store cookies: n = 12, mean = 7, s = 1. Test your theory, using the 5% level of significance. Be sure to identify the null and alternative hypotheses, identify the appropriate distribution and indicate the tail(s) for the critical region, identify and calculate the test statistic, determine the P-value, make a decision and state a conclusion.
To test your theory and determine if there is a significant difference in the number of chocolate chips between Mom's homemade cookies and the store brand cookies, you can conduct a hypothesis test.
First, let's define our null and alternative hypotheses:
Null hypothesis (H0): There is no difference in the mean number of chocolate chips between Mom's cookies and the store brand cookies.
Alternative hypothesis (Ha): Mom's cookies have a greater mean number of chocolate chips than the store brand cookies.
Next, let's identify the appropriate distribution to use for this hypothesis test. Since we are comparing two sample means and do not know the population standard deviation, we will use the t-distribution.
Since we believe that Mom's cookies have more chips, we are interested in testing if the mean of Mom's cookies is greater than the mean of the store brand cookies. Therefore, this is a one-tailed test.
Now, let's calculate the test statistic. We'll use the formula for calculating the t-statistic for independent samples:
t = (x1 - x2) / sqrt((s1^2 / n1) + (s2^2 / n2))
Where:
x1 = mean number of chocolate chips in Mom's cookies
x2 = mean number of chocolate chips in store brand cookies
s1 = standard deviation of chocolate chips in Mom's cookies
s2 = standard deviation of chocolate chips in store brand cookies
n1 = number of observations in Mom's cookies sample
n2 = number of observations in store brand cookies sample
Plugging in the values, we have:
x1 = 9
x2 = 7
s1 = 3
s2 = 1
n1 = 12
n2 = 12
t = (9 - 7) / sqrt((3^2 / 12) + (1^2 / 12))
t = 2 / sqrt(0.75 + 0.0833)
t = 2 / sqrt(0.8333)
t ≈ 2.68
Next, we need to determine the P-value associated with the test statistic. Since the alternative hypothesis states that Mom's cookies have a greater mean number of chocolate chips, we need to find the P-value for a t-distribution with (n1 + n2 - 2) degrees of freedom, where n1 and n2 are the sample sizes of the two groups.
To find the P-value, we can use software, tables, or a t-distribution calculator. The P-value is the probability of observing a t-value greater than or equal to 2.68 (for a one-tailed test).
Assuming the P-value is less than the specified significance level (5%), we can make a decision. If the P-value is less than 0.05, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Based on your data and the specified significance level, perform the calculation for the P-value and make a decision. Then, state the conclusion of the hypothesis test.