Webcalc Ii

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  • Topic: Normal distribution, Standard deviation, Statistics
  • Pages : 7 (1569 words )
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  • Published : June 22, 2012
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Z test &One sample t-test

1. A researcher is interested in whether students who attend private high schools have higher average SAT Scores than students in the general population. A random sample of 90 students at a private high school is tested and and a mean SAT score of 1030 is obtained. The average score for public high school student is 1000 (σ= 200).

a. Is this a one- or two tailed test?

Key word “whether”.This is a two-tailed hypothesis test. Researcher wants to know “whether” the groups being compared differ, but does not predict the direction of the difference.

The mean of the sample will be different from or unequal to the mean of the general population.

b. What are H0 and Ha for this study?

The researcher predicts a difference in SAT scores between private high schools and those in public high schools, but the direction of the difference is not predicted. Those in private high schools would be expected to have either higher or lower SAT scores but not the same SAT score as the public high schools.

The statistical notation for this two-tailed test is

H0:µ0 = µ1, or µPrivate High School = µ Public High School
Ha:µ0 = µ1, or µPrivate High School ≠ µ Public High School

c. Compute Zobt

The mean SAT score of the public high school (µ) is 1000, with a standard deviation (σ) of 200; for private high school in the sample (N=90), mean SAT score is 1030.

σx= σ/Square Root of N = 1000/Square Root of 200 = 1000/14.14 = 70.72
z= X - µ/σx= 1030 – 1000/70.72 = 30/70.72 = .42

Zobt= .42

[All calculations were done by using hand and/or 10 digit calculator.]

d. What is the Z critical value (Z cv ) using a 0.05 alpha level?

According to table A.2 0.05 alpha level is 1.96

e. Should H0 be rejected? What should the researcher conclude?

Zobthas to be as large or larger than Z cvto reject H0
Zobt= .42
Z cv= 1.96

Zobt< Z cv, therefore H0 should be rejected.

The researcher should conclude that private high schools do not havehigher average SAT scores.

f. Determine the 95 % confidence interval for the population mean, based on the sample mean.

The mean SAT score of the public high school (µ) is unknown, with a standard deviation (σ) of 200; for private high school in the sample (N=90), mean SAT score is (1030).

σx = 200/Square Root of 90 = 200/9.5 = 21.05

We can calculate the 95% confidence interval using the following formula CI = X ± z(σx)

CI=1030±1.96(21.05)
=1030±41.26
=989 – 1071

Thus, the 95% confidence interval ranges from 989 – 1071. We could conclude, based on this calculation, that we are 95% confident that the population mean lies within this interval.

[All calculations were done by using hand and/or 10 digit calculator.]

2. A researcher hypothesized that the pulse rates of long-distance athletes differ from those of other athletes. He believed that the runners’ pulses would be slower. He obtained a random sample of 10 long-distance runners. He measured their resting pulses. Their pulses were 45, 45, 64, 50, 58, 49, 47, 55, 50, 52 beats per minute. The average resting pulse of athletes in the general population is normally distributed with a pulse rate of 60 beats per minute. a. What statistical test should be used to analyze the data?

The Student t distribution test should be used to analyze this data. The t test is a means of determining the number of standard deviation units a score is from the mean (µ) of a distribution. Population variance is not known and data does not fit the standard normal distribution.

b. Is this a one- or two- tailed test?

Key word: believed. This is a one-tailed hypothesis test. Researcher is conducting a test in support of his belief that endurance runners’ have slower pulses.

The mean of the sample will be greater than the mean of the population, or the mean of the sample...
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