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.
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.
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.
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 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.
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.
Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.
