Estimate the number of times an average person's heart beats in a lifetime. assume the average heart rate is 69 beats/min and a
2 answers:
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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!