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Someone plz plz help with this Statistics Intro to R programming question!!!

Here are the examples and follow by my question!!

Thank you so much!! I appreciate it

Question Type 1: If possible, calculate the 90% confidence intervals for the temperature it takes for crickets to chirp 15 ch

0 8 70 75 80 85 90 Degrees Farenheit Conclusion of Data Analysis Based on the probability plot there does not seem to be any

Results: From the Console copy and paste the output generated from the previous command) One Sample t-test data: CricketssTem

uestion type 2: The researcher believes that the mean temperature has changed from 76°F. Test his claim at a 0.05 significanc

!!!!My question!!!!

Question 2: A researcher is looking at the height of a new hybrid strawberry bush. A random sample of 17 bushes heights was r

Based on the output of R, what is your p-value (if possible, based on conclusion in part b of question 1) e. f. Determine whe

Strawberr 11.4 10.6 11.3 11.6 10.9 10.8 11.1 10.8 11.4 11.4 11.2 10.5 10.7 10.9 10.7

Question Type 1: If possible, calculate the 90% confidence intervals for the temperature it takes for crickets to chirp 15 chirps per second. Code (you must copy and paste your code like below in blue color): # Reading in the data Crickets-read.table(C:/Desktop/CricketChirpsvsTemperature.csv', header TRUE, #View Data View Crickets) #Data analysis to see if sample is normal so that we can reasonably assume #data comes from a normal population anoumiCrickets$Tempmain "Temperature") aalinelickets$teme col "blue") main "Temperature" horizontal TRUE xlab- "Degrees Eareabeit col "blue") Results on next page (you must copy and paste your graphs-copy the two graphs from R and put it below from the lower right plots in R-Studeo -Use Export dropdown, Copy to clipboard, to copy graplh then paste it below
0 8 70 75 80 85 90 Degrees Farenheit Conclusion of Data Analysis Based on the probability plot there does not seem to be any deviation from normality. Further, the boxplot indicates there are no outliers. Therefore, it is reasonable to assume that this sample of temperatures came from a population that is normally distributed. 90% confidence interval: Code: #Fastest way to get a confidence interval (90% CI in this example), but you will get extra information, disregard cantlevel 0.90)
Results: From the Console copy and paste the output generated from the previous command) One Sample t-test data: CricketssTeme. t46.217, df 14, p-value
0 0
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Answer #1

Solution(2)

(1)

Rcode:

strawberry <- c(11.4,10.6,11.3,11.6,11,10.9,10.8,11.1,11,10.8,11.4,11.4,11.2,10.5,10.7,10.9,10.7)
print(strawberry)
t.test(strawberry,conf.level = 0.90)

90 percent confidence interval lies in between
10.88106 and 11.15423

Solution2a:

qqnorm(strawberry,main="new hybrid strawberry bush")
qqline(strawberry,col="green")
boxplot(strawberry,main="new hybrid strawberry bush",horizontal=TRUE,xlab="In inches",col="green")

Solution2b:

new hybrid strawberry bush E.N 03 -2 0 2 Theoretical Quantiles

new hybrid strawberry bush 11.6 11.4 11.2 11.0 10.8 10.6 In inches

Solution2c:

no outliers found from boxplot .

Form qqplot data does not deviate from normal distribution.

Therefore the sample is taken from population that is normally distributed.

Solution2d:

strawberry <- c(11.4,10.6,11.3,11.6,11,10.9,10.8,11.1,11,10.8,11.4,11.4,11.2,10.5,10.7,10.9,10.7)
print(strawberry)
t.test(strawberry,conf.level = 0.90)

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