A grocery store’s receipts show that Sunday customer purchases have a skewed distribution with a mean of $32 and a standard devi
1 answer:
Answer:
a) This is because the Sunday customer purchases is skewed. b) For a highly skewed distribution, it is not possible but if it is slightly skewed, it is possible. c) approximately 0.0023
Step-by-step explanation:
a) As stated in the question, the Sunday customer purchases is skewed, thus we cannot use the normal model to estimate the probability that the next Sunday customer will spend at least $40.
b) Using a sample of 10 Sunday customers won't be large enough to apply the Normal model for sampling. However, if it is just slightly skewed, the sample may be large enough for the analysis while if it is highly skewed it would be impossible to calculate the probability.
c) Using randomization principle, it can be assumed that 50 Sunday customer purchases can be considered a representative sample for all Sunday customer purchases. Using a 10% condition for the computation:
μ = μ = $32
σ = 20/
= 2.828
Then, the z-score is:
Thus:
P (z ≥ 2.83) = normalcdf (2.83, E99, 0.1) = 0.0023
Therefore, the probability that the average of the Sunday customer purchases is at least $40 is 0.0023
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Answer:
Two, one for the 14 responses (number of visits) by the adults who fear going to the dentist and one for the 31 responses (number of visits) by the adults who do not fear going to the dentist.
Step-by-step explanation:
Hello!
1)
You want to test if the average visits to the dentist of people who fear to visit it are greater than the average visits of people that don't fear it.
In this case, the statistic to use is a pooled Student t-test. The reason I've to choose this test is that one of your sample sizes is small (n₁= 14) and the t-test is more accurate for small samples. Even if the second sample is greater than 30, if both variables are normally distributed, the pooled t-test is the one to use.
H₀: μ₁ = μ₂
H₁: μ₁ > μ₂
α: 0.10
t=<u> (X₁[bar]-X₂[bar]) - (μ₁ - μ₂)</u> ~ t
Sₐ√(1/n₁+1/n₂)
Where
X₁[bar] and X₂[bar] are the sample means of both groups
Sₐ is the pooled standard deviation
This is a one-tailed test, you will reject the null hypothesis to big numbers of t. Remember: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis), and in this case, is also one-tailed.
P(t ≥ t
) = 1 - P(t
< t
)
Where t is the value of the calculated statistic.
Since you didn't copy the data of both samples, I cannot calculate it.
2)
Well there was one sample taken and separated in two following the criteria "fears the dentist" and "doesn't fear the dentist" making two different samples, so this is a test for two independent samples. To check if both variables are normally distributed you need to make two QQplots.
I hope it helps!