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Answer:
1. M=108
2. μ=110
3. In the explanation.
4. Test statistic t = -1.05
5. P-value = 0.1597
Step-by-step explanation:
The question is incomplete: to solve this problem, we need the sample information: size, mean and standard deviation.
We will assume a sample size of 10 matches, a sample mean of 108 points and a sample standard deviation of 6 points.
1. The mean points is the sample points and has a value of 108 points.
2. The null hypothesis is H0: μ=110, meaning that the mean score is not significantly less from 110 points.
3. This is a hypothesis test for the population mean.
The claim is that the mean score is significantly less than 110.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample has a size n=10.
The sample mean is M=108.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.
The estimated standard error of the mean is computed using the formula:
4. Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size are:
5. This test is a left-tailed test, with 9 degrees of freedom and t=-1.05, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.1597) is bigger than the significance level (0.05), the effect is not significant.
The null hypothesis failed to be rejected.
There is not enough evidence to support the claim that the mean score is significantly less than 110.
Answer:
The lowest score on the final exam that would qualify a student for an A is 80.
The lowest score on the final exam that would qualify a student for a B is 74.68.
The lowest score on the final exam that would qualify a student for a C is 67.33.
The lowest score on the final exam that would qualify a student for a D is 62.
Step-by-step explanation:
When the distribution is normal, we use 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.
Mean of 71 and a standard deviation of 7.
This means that
Grades are assigned such that the top 10% receive A's, the next 20% received B's, the middle 40% receive C's, the next 20% receive D's, and the bottom 10% receive F's.
This means that:
90th percentile and above: A
70th percentile and below 90th: B
30th percentile to the 70th percentile: C
10th percentile to the 30th: D
Lowest score for an A:
Top 10% receive A, which means that the lowest score that would qualify a student for an A is the 100 - 10 = 90th percentile, which is X when Z has a pvalue of 0.9, so X when Z = 1.28.
The lowest score on the final exam that would qualify a student for an A is 80.
Lowest score for a B:
70th percentile, which is X when Z has a pvalue of 0.7, so X when Z = 0.525.
The lowest score on the final exam that would qualify a student for a B is 74.68.
Lowest score for a C:
30th percentile, which is X when Z has a pvalue of 0.3, so X when Z = -0.525.
The lowest score on the final exam that would qualify a student for a C is 67.33.
Lowest score for a D:
10th percentile, which is X when Z has a pvalue of 0.1, so X when Z = -1.28.
The lowest score on the final exam that would qualify a student for a D is 62.