Suppose the mean number of push-ups done in 1 minute by elementary school childr

Question

Suppose the mean number of push-ups done in 1 minute by elementary school children as measured in 2010 was 9.8. However, in the Flenzheim school district this year , a sample of n = 15 students in the 4th grade was taken and the average number of push-ups performed in 1 minute, Mean was 10.1 with a standard deviation, s, of 0.8. Run a test of hypothesis to determine if the mean number of push-ups done in 1 minute by elementary school children has increased. Assume the population is normally distributed. Use = 0.05 as the level of significance. (A) Set up the null and alternative hypotheses. H0: ____ H1: ____ This is a LEFT/RIGHT/TWO-TAILED TEST (Circle one) (B) Write down the level of significance: = __________. (C) Which distribution must I use in running the test? TI 83/84 calculator Select the proper test based on your answer to (C) above. Inpt: (Make sure “Stats” is highlighted) _0 = _______________ x^- = _______________ S_x = _______________ n = ______________ : (Select one: _0 , < _0 >, _0) Color: (Just arrow down) Calculate: What is the p-value? ___________________ Compare: If the p-value < , REJECT the null hypothesis. Is the p-value < ? ___________. Then we REJECT/ DO NOT REJECT the null hypothesis (circle one).

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: < 9.8

Alternative hypothesis: > 9.8

Note that these hypotheses constitute a right-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.207

DF = n - 1 = 15 - 1

D.F = 14

t = (x - ) / SE

t = 1.45

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of 1.45. We use the t Distribution Calculator to find P(t > 1.45) = 0.0735

Thus the P-value in this analysis is 0.0735

Interpret results. Since the P-value (0.0735) is greater than the significance level (0.05), we cannot reject the null hypothesis.

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