Question

To be completed with code from R Studio. No additional information provided.

In AU supermarkets, the average weight of a banana is 119.9 grams. An agricultural scientist buys bananas from a supermarket. Their weight, in grams, is as follows:

c(113.9, 90.5, 93.9, 113.4, 101, 76.4)

She suspects that this sample of bananas is lighter than average and wonders if this supermarket is selling bananas that are lighter than the AU average.

(a) State a sensible null hypothesis

(b) State the precise definition of p-value and explain what “more extreme” means in this context

(c) Is a one-sided or two-sided test needed? justify

(d) Perform a student t-test using R and interpret

(e) Perform a Z test and account for any differences you find

Answer 1

Solution-:

Given data: , n=6

113.9.90.5.93.9.113.4.101.76.4

By using RStudio:

> # Hypothesis
> #(a) Null Hypothesis Ho: The average weight of a bananasis 119.
> #Vs
> # Alternative Hypothesis H1: The average weight of a bananas is lighter than 119.
> #(d) mu=119.9
> x=c(113.9,90.5,93.9,113.4,101,76.4);x
[1] 113.9 90.5 93.9 113.4 101.0 76.4
> t.test(x,mu=119.9,conf.level=0.95,alt="l")

   One Sample t-test

data: x
t = -3.6913, df = 5, p-value = 0.007064
alternative hypothesis: true mean is less than 119.9
95 percent confidence interval:
-Inf 110.0384
sample estimates:
mean of x
98.18333

(c) This is one sided, beacuse alternative hypothesis is left tailed.

(b) The P-value is the probability of a larger effect than what was observed.

Here, P-value= 0.007064 is less than level of significance =0.05 so we reject the null hypothesis.

(e) Here, $\sigma$ is unkown so  we can not perform or apply Z- test.

RStudio Code:

# Hypothesis
#(a) Null Hypothesis Ho: The average weight of a bananasis 119.
#Vs
# Alternative Hypothesis H1: The average weight of a bananas is lighter than 119.
#(d) mu=119.9
x=c(113.9,90.5,93.9,113.4,101,76.4);x
t.test(x,mu=119.9,conf.level=0.95,alt="l")