Question

RStudio / null & alternate hypothesis testing

- Question -

The population mean µ of the full-body lengths (cm) is taken of an undisclosed species today (2020). Jack decides to compare this undisclosed species full-body lengths mean from 2020 with the mean from a decade ago. The mean full-body lengths from a decade ago is 50 cm.
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(Other information):
Sample size = 38.
Standard deviation = 4.5
Full-body lengths mean = 43.34cm
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(a) Hypothesis test, has the mean (µ) changed from 50 cm?
(b) What's the null and alternative hypothesis?
(c) What's an appropriate test statistic?
(d) What's the observed value in the sample?
(e) What's the sampling distribution for this statistic?
(f) P-value?

Answer 1

Let $\mu$ denotes the true mean full-body length.

Let u denotes the population mean. Hypotheses : To test null hypothesis Ho : =50 against alternative hypothesis H1 : #50 Let z denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, s 4.50 T = 43.3400, s2 = 20.2500, s = 4.5000, n = 38, standard error of mean = = 0.7300 38 The test statistic can be written as : Vn(– 50) t= S which under H, follows at distribution with n - 1 df. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tobs > to.025,n–1 = 2.026192 0.4 0.3 0.2 0.1 Mim 0.0 t Now, Value of the test statistic : The value of the test statistic tobs 38 (43.3400 – 50) 4.5000 -9.123333 -9.12 associated degrees of freedom = 37 The critical value =Łt 0.05 = Eto.025,37 = +2.026192 +2.0262 2,9-1 P-value = P(|tn-11 > tobs) = 2 * P(t37 <-9.123333) = 2 * 0.000000 =0.000000 0.0000 Conclusion : Since p-value < 0.05 and tobs | > tcritical = 2.026192, so we reject H, at 0.05 level of significance Hence, we can conclude that the the population mean is significantly different from 50.