you gave me this answer in last question, (copied below) but I see why I wasn't finding the correct p-value. I don't know how to get this from excel or in the z table. My z table gives me answeres to the left of z. and when I put my excel formula in which is =norm.s.dist(1.99, true) I get same answer as in z table to the left. A-how do I get answer for p-value when mu is greater than 85 like below problem? B- Also, how do I get answer for P-value when its a two tail test and mu differs? Step 1 Given data of glucose reading for 8 weeks92 87 82 107 99 109 85 89The sample x-bar is 93.8 . x is normal distribution with standard deviation = 12.5 Given that the mean glucose levelfor horses is µ = 85 mg/100glclaim is that to see if the gentle Ben have glucose level higher than 85 at 0.05 level of significance. Null hypothesis : µ = 85Alternative hypothesis : µ > 85. As x is normally distributed we use one sample Z test. So the test statistic Z is calculate dby using the below mentioned formula and by substituting values we get the test statistic as 1.99 (after rounding to two decimals). help_outlinefullscreen Step 2 With reference to the Z table we can see that the area to the right of 1.99 is 0.0233. As this is one tail test. The p value is nothing but the the proportion of data to right of test statistic p value = P( X > 1.99) = 0.0233.
Inverse Normal Distribution
The method used for finding the corresponding z-critical value in a normal distribution using the known probability is said to be an inverse normal distribution. The inverse normal distribution is a continuous probability distribution with a family of two parameters.
Mean, Median, Mode
It is a descriptive summary of a data set. It can be defined by using some of the measures. The central tendencies do not provide information regarding individual data from the dataset. However, they give a summary of the data set. The central tendency or measure of central tendency is a central or typical value for a probability distribution.
Z-Scores
A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow comparison between two scores that belong to different normal distributions.
Given data of glucose reading for 8 weeks
92 87 82 107 99 109 85 89
The sample x-bar is 93.8 .
x is normal distribution with standard deviation = 12.5
Given that the mean glucose levelfor horses is µ = 85 mg/100gl
claim is that to see if the gentle Ben have glucose level higher than 85 at 0.05 level of significance.
Null hypothesis : µ = 85
Alternative hypothesis : µ > 85.
As x is
With reference to the Z table we can see that the area to the right of 1.99 is 0.0233. As this is one tail test. The p value is nothing but the the proportion of data to right of test statistic p value = P( X > 1.99) = 0.0233.
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