According to the Centers for Disease Control, 33 of American adults ages 20 and over suffer from hypertension A sample of 102 adults were recruited to participate in relaxation training in hopes of lowering the incidence of hypertension At the end of a year, the adults are re evaluated and the proportion who suffer from hypertension is recorded Completely describe the sampling distribution of the sample proportion of adults who suffer from hypertension when samples of size 102 are selected Round any calculated values to four decimal places mean p p standard deviation p p (round to 4 decimal places) np and n(1 p) so the distribution of p p is Select an answer Normally distributed Not normally distributed According to the Empirical Rule, we expect 68 of all samples of 102 adults to result in a sample proportion betweenandwith hypertension Round your answers to 4 decimal places ( Hint Draw the normal model based on the description in part a ) Suppose the sample of 102 adults participated in the relaxation training and returns a proportion of 0 214 with hypertension Does this sample proportion provide any support for the conjecture made The sample proportion, 0 214, is between 2 and 3 standard deviations below the mean This provides strong support for the conjecture that relaxation training lowered the occurrence because it is unlikely to occur by chance alone The sample proportion, 0 214, is more than 3 standard deviations below the mean This provides very strong support for the conjecture that relaxation training lowered the occurrence because it is highly unlikely to occur by chance alone The sample proportion, 0 214, is less than 1 standard deviation below the conjectured value This would not provide strong support to the conjecture that relaxation training lowered the occurrence of hypertension The sample proportion, 0 214, is between 1 and 2 standard deviations below the mean This provides some support for the conjecture that relaxation training lowered the occurrence, although it is not overwhelming

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According to the Centers for Disease Control, 33 % of American adults ages 20 and over suffer from hypertension. A sample of 102 adults were

According to the Centers for Disease Control, 33 % of American adults ages 20 and over suffer from hypertension. A sample of 102 adults were recruited to participate in relaxation training in hopes of lowering the incidence of hypertension. At the end of a year, the adults are re-evaluated and the proportion who suffer from hypertension is recorded.

Completely describe the sampling distribution of the sample proportion of adults who suffer from hypertension when samples of size 102 are selected.Round any calculated values to four decimal places.

mean:
p
p^
=
standard deviation:
p
p^
=(round to 4 decimal places)
np =and n(1 - p) =so the distribution of
p
p^
is:Select an answer
Normally distributed
Not normally distributed

According to the Empirical Rule, we expect 68 % of all samples of 102 adults to result in a sample proportion betweenandwith hypertension. Round your answers to 4 decimal places. (Hint:Draw the normal model based on the description in part a.)
Suppose the sample of 102 adults participated in the relaxation training and returns a proportion of 0.214 with hypertension. Does this sample proportion provide any support for the conjecture made?

The sample proportion, 0.214, is between 2 and 3 standard deviations below the mean. This provides strong support for the conjecture that relaxation training lowered the occurrence because it is unlikely to occur by chance alone.
The sample proportion, 0.214, is more than 3 standard deviations below the mean. This provides very strong support for the conjecture that relaxation training lowered the occurrence because it is highly unlikely to occur by chance alone.
The sample proportion, 0.214, is less than 1 standard deviation below the conjectured value. This would not provide strong support to the conjecture that relaxation training lowered the occurrence of hypertension.
The sample proportion, 0.214, is between 1 and 2 standard deviations below the mean. This provides some support for the conjecture that relaxation training lowered the occurrence, although it is not overwhelming.