We determine that the true average calorie content as estimated in the sampled population surpasses the actual calorie content. Step-by-step explanation: An article discussed a pilot study where each of the 58 participants was asked to estimate the calorie count of a 12 oz beer known to have 153 calories. The observed sample mean of calorie estimation was 193, with a sample standard deviation of 88. Let

=
true average estimated calorie level within the sampled population. Thus, Null Hypothesis, :
153 calories {indicating that the true average estimated calorie content does not exceed the actual amount}. Alternative Hypothesis,
:
> 153 calories {indicating the true average estimation exceeds the actual}. The appropriate test statistic would be a one-sample t-test statistic, as we lack knowledge of the population standard deviation; Test Statistic = ~t = 
where, sample mean estimated calorie level = 193 calories, s = sample standard deviation = 88, and n = sample size = 58. Therefore, the test statistic = ~t = 3.462. The t-table indicates a critical value of 1.6725 for 57 degrees of freedom at a 0.05 significance level. Since our test statistic of 3.462 > 1.6725, we have sufficient evidence to reject the null hypothesis; thus, affirming that the true average estimated calorie content in the sampled population exceeds the real content.
Answer:
Find below:
Step-by-step explanation:
To determine this, we will either calculate the total cost of acquiring 40 bouquets at $2.50 each or find the single bouquet’s cost at $120.
Cost of one in pack of 40 priced at $120.
120 divided by 40 equals $3
Now, we notice that $3>$2.50
This indicates Kendra has made an error by purchasing the 40 bouquet pack at $120
Hope this helps.
Good Luck
Response:
Option B is correct
Step-by-step breakdown:

(One-tailed test at a significance level of 5%)
n = 16 and x̄ = 1.97
s = 0.1
Standard error of the mean = 
Difference in means = 
t statistic = Difference in means/se = -1.2
degrees of freedom = 16 - 1 = 15
p-value = 0.124375
(B) do not reject the null hypothesis since the test statistic (-1.2) is > the critical value (-1.7531).
Answer:
Yes, Bill surpassed his target by £2,200.
Step-by-step explanation:
The cost for the laptops totals = 50 x 400
= 20,000
He sold 40 laptops with a profit margin of 30%
Calculating that gives (40 x 400) x 1.30
= 20,800
He sold 10 laptops with a profit of 15%
Calculating that gives (15 x 400) x 1.15
= 6900
Thus, the overall sales revenue totals
= 20800 + 6900 = 27,700