For questions #8-11: We are interested in exploring the average amount spent on marketing in the last quarter by retail outlets in the US, which we assume are normally distributed. An article in the Wall Street Journal reported that the average amount spent on marketing in the last quarter by retail outlets in the US was $551,715.69.Suppose the firm you are working for randomly samples 800 retail outlets and finds that the sample average is $547,195 with a sample standard deviation of $82,681.

8.    Your partner Bob's firm believed, before gathering data, that the true average amount spent on marketing is different than the amount reported by the Wall Street Journal. Is there enough evidence to find that the partner firm is correct?

 

Answer:

Claim: The true average amount spent on marketing is different than the amount reported by the Wall Street Journal. 

From the above claim, the null hypothesis ?(H0?)? and the alternative hypothesis ?(H1?)? are as follows:

 

?H0?:µ=551715.69?

 

?H1?:µ? ≠ 551715.69

 

Answer: The value of test statistics t = -1.5465.

 

Answer: The p-value is 0.1224.

 

Decision: We fail to reject the null hypothesis.

 

Conclusion: At 5% level of significance, there is insufficient evidence to support the claim that the true average amount spent on marketing is different than the amount reported by the Wall Street Journal.

 

Interpretation: The data suggest that the average amount spent on marketing in the last quarter by retail outlets in the US was $551,715.69.

Step-by-step explanation

Answer:

 

Claim: The true average amount spent on marketing is different than the amount reported by the Wall Street Journal. 

 

From the above claim, the null hypothesis ?(H0?)? and the alternative hypothesis ?(H1?)? are as follows:

 

?H0?:µ=551715.69?

 

?H1?:µ? ≠ 551715.69

 

Assumption: Here population standard deviation is not given, the sample size is 800 (n = 800 > 30) which is large enough for the Central Limit Theorem to apply. Therefore, the assumptions are satisfied. So we can assume that the sample comes from the normal population, so we can use one sample t-test to test the above hypothesis. From the alternate hypothesis we will use the two tailed one sample t-test.

 

The formula of t-test statistic is as follows:

 

?t=s/n?x?−μ?...(1)?

 

Given that,

Sample mean = ?x?? = 547195, 

Sample size = n = 800,

Sample standard deviation = s = 82681 and

 

Plugging these values in equation (1), we get

 

?t=82681/800?547195−551715.69??

?t=2923.21−4520.69?=−1.5464789?

t = -1.5465 (Rounded to four decimal places)

Answer: The value of test statistics t = -1.5465.

Let's find the p-value:

From the alternative hypothesis, this is the two tailed t-test.

So, p-value = 2*P(T > |t|)

Here, t = -1.5465

So, |t| = 1.5465

df = n - 1 = 800 - 1 = 799

Here, we need to use technology as excel to find the p-value.

The general command to find two tailed probability from the t-value in excel is "=TDIST(|t|, df, 2)".

So, p-value = "=TDIST(1.5465, 799, 2)" = 0.1224 (Rounded to four decimal places).

Answer: The p-value is 0.1224.

Let us assume the level of significance = α = 0.05

Decision rule:

1) If p-value < level of significance (α) then we reject null hypothesis

2) If p-value > level of significance (α) then we fail to reject null hypothesis.

Since p-value = 0.1224 > 0.05 = α , so here we used the second rule that is we fail to reject the null hypothesis.

Decision: We fail to reject the null hypothesis.

Conclusion: At 5% level of significance, there is insufficient evidence to support the claim that the true average amount spent on marketing is different than the amount reported by the Wall Street Journal.

Interpretation: The data suggest that the average amount spent on marketing in the last quarter by retail outlets in the US was $551,715.69.