According to the Centers for Disease Control, 31 8 of American adults ages 20 and over suffer from hypertension A sample of 108 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 108 are selected Round any calculated values to four decimal places Mean pp Standard deviation p (round to 4 decimal places) Shape np and n(1 p) so the distribution ofpp is Select an answer Not normally distributed Normally distributed According to the Empirical Rule, we expect 95 of all samples of 108 adults to result in a sample proportion between and with hypertension Round your answers to 4 decimal places ( Hint Draw the normal model based on the description in part a ) Suppose the sample of 108 adults participated in the relaxation training and returns a proportion ofp 0 1921 with hypertension Does this sample proportion provide any support for the conjecture made a) The sample proportion, 0 1921, 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 b) The sample proportion, 0 1921, 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 c) The sample proportion, 0 1921, 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 d) The sample proportion, 0 1921, 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

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According to the Centers for Disease Control, 31.8% of American adults ages 20 and over suffer from hypertension. A sample of 108 adults were recruited

According to the Centers for Disease Control, 31.8% of American adults ages 20 and over suffer from hypertension. A sample of 108 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 108 are selected.Round any calculated values to four decimal places.
- Mean: pp^= ________?
- Standard deviation:p^=_________? (round to 4 decimal places)
- Shape: np = _________?and n(1 - p) = _________? so the distribution ofpp^is: Select an answer Not normally distributed Normally distributed
According to the Empirical Rule, we expect 95% of all samples of 108 adults to result in a sample proportion between ________? and _______? with hypertension. Round your answers to 4 decimal places. (Hint:Draw the normal model based on the description in part a.)
Suppose the sample of 108 adults participated in the relaxation training and returns a proportion ofp^=0.1921 with hypertension. Does this sample proportion provide any support for the conjecture made?

a) The sample proportion, 0.1921, 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.

b) The sample proportion, 0.1921, 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.

c) The sample proportion, 0.1921, 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.

d) The sample proportion, 0.1921, 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.