Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is less than 1
1 answer:
Answer:
0.8749
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
The probability that Z is less than 1.15 is:
This is the pvalue of Z = 1.15, which is 0.8749.
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Answer:
The p value is higher than any significance level given for example , so then we can conclude that we FAIL to reject the null hypothesis. So we don't have enough evidence to conclude that the mean reading is greater for the auscultatory method at 5% or 10% of significance.
Step-by-step explanation:
A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations we can use it.
Let put some notation
x=Auscultatory method , y = Oscillatory method
x: 74 86 84 79 70 78 79 70
y: 70 85 90 110 71 80 69 74
The system of hypothesis for this case are:
Null hypothesis:
Alternative hypothesis:
The first step is define the difference , that is given so we have:
d: 6.6, 4.2, -5.5, -3.1, 9.3, -3.9
The second step is calculate the mean difference
The third step would be calculate the standard deviation for the differences, and we got:
The fourth step is calculate the statistic given by :
The next step is calculate the degrees of freedom given by:
Now we can calculate the p value, since we have a right tailed test the p value is given by:
The p value is higher than any significance level given for example , so then we can conclude that we FAIL to reject the null hypothesis. So we don't have enough evidence to conclude that the mean reading is greater for the auscultatory method at 5% or 10% of significance.