A study was made on the amount of converted sugar in a certain process at variou

Question

A study was made on the amount of converted sugar in a certain process at various temperatures. The data were coded and recorded as follows:
Temperature, x: 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0,

6. Test = 0 against > 0 at = 0.01. 7. Compute a two-sided 95% confidence interval for the mean response when x = 1.6. 8. Compute a two-sided 95% prediction interval for a single predicted value of the amount of converted sugar when x = 1.6. 9. Calculate the correlation coefficient r. 10. Test the hypothesis that -0 against the alternative that 0 at the 0.05 level of significance. What is your conclusion?

Explanation / Answer

we shall do the analysis in R ,

The complete R snippet to analyse this as follows

Temperature <- c( 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0)
ConvertedSugar <- c(8.1, 7.8, 8.5, 9.8, 9.5, 8.9, 8.6, 10.2, 9.3, 9.2, 10.5)

## fit the model
fit <- lm(ConvertedSugar~Temperature)
summary(fit)

## confidence interval
confint(fit,Temperature=1.6,level = 0.95)

# prediction interval
predict(fit,newdata=data.frame(Temperature=1.6),interval = "predict")

## correlation
cor(Temperature,ConvertedSugar)

## correlation test
cor.test(Temperature,ConvertedSugar)

The result is

summary(fit)

Call:
lm(formula = ConvertedSugar ~ Temperature)

Residuals:
Min 1Q Median 3Q Max
-0.7082 -0.4868 -0.1227 0.5109 1.0346

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 6.4136 0.9246 6.936 6.79e-05 ***
Temperature 1.8091 0.6032 2.999 0.015 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.6326 on 9 degrees of freedom
Multiple R-squared: 0.4999,   Adjusted R-squared: 0.4443
F-statistic: 8.996 on 1 and 9 DF, p-value: 0.01497

> confint(fit,Temperature=1.6,level = 0.95)
2.5 % 97.5 %
(Intercept) 4.3219598 8.505313
Temperature 0.4446316 3.173550
> # prediction interval
> predict(fit,newdata=data.frame(Temperature=1.6),interval = "predict")
fit lwr upr
1 9.308182 7.807277 10.80909

> cor(Temperature,ConvertedSugar)
[1] 0.7070264

> ## correlation test
> cor.test(Temperature,ConvertedSugar)

   Pearson's product-moment correlation

data: Temperature and ConvertedSugar
t = 2.9993, df = 9, p-value = 0.01497 , as the p value is less than 0.05 , hence we reject the null hypothesis and conclude that the correlation is signficant
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
0.1860680 0.9176857
sample estimates:
cor
0.7070264

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