Question

The following table shows voltage readings collected from solar panel on three different types of day. Use One-way ANOVA method to answer following questions. Do this problem using StatDisk. Use a - 0.05. You must attach StatDisk output. Round values to three significant digits if necessary Sunny davs Cloudy daysRainy days 13.5 13.0 13.2 13.9 13.8 12.7 12.5 12.6 12.7 13.0 13.0 12.1 12.2 12.3 11.9 11.6 12.2

a) What is null hypothesis in terms of voltage readings?

b) Find the variance between treatments. (MSbetween)

c) Find the variance within treatments. (MSwithin)

d) Compute the F ratio ( F test statistic, Ft.s.)

e) Identify the numerator degree of freedom and denominator degree of freedom.

f) Identify the F critical value F c.v.

g) What can you conclude about the claim that the three different types of day

populations have the same mean voltage readings? Do you reject null hypothesis? Explain your

answer in terms of test statistic and critical value on F -distribution graph.

h) If you concluded that there is a difference in voltage readings among three different

types of day, use Tukey’s HSD Test to determine which reading is different.

Answer 1

Null hypothesis: That is, voltage readings collected from solar panel on three different types of day are same. EXCEL output:

Anova: Single Factor SUMMARY Groups Count Sum Average Variance 81.4 13.56667 0.162667 12.75 0.043 0.067 Sunny days Cloudy days Rainy days 72.3 12.05 ANOVA Source of Variation Between Groups Within Groups MS P-value F crit 6.914444 2 3.457222 38.0379 1.33E-06 3.68232 15 0.090889 1.363333 Total 8.277778 17 The value of variance between treatments is MSBn«es-3.4572 Berween The value of variance within treatments is MS0.0909

The value of test statistic is F 38.0379 Numerator degrees of freedom Denominator degrees of freedom 15 The critical value is Fof -3.6823 Since Fcai 38.0379)>F3.6823), then reject the null hypothesis. Therefore, the three different types of day populations have different mean voltage readings.

Tukey simultaneous comparison t-values (d.f. 15) Rainy days Cloudy days 12.75 Sunny days 13.57 Rainy days 12.05 Cloudy days 12.75 Sunny days 13.57 4.02 8.71 4.69 critical values for experimentwise error rate 0.05 0.01 2.60 3.42 Since Fi 4.02)> Fct2.60), then reject the null hypothesis Therefore, the means of Cloudy days and Rainy days are different Since Fai8.7F-2.60), then reject the null hypothesis Therefore, the means of Sunny days and Rainy days are different Since Faic4.69F -2.60), then reject the null hypothesis Therefore, the means of Sunny days and Cloudy days are different