Question

A least squares adjustment is computed twice on a data set. When the data is minimally constrained with 20 degrees of freedom, a reference variance of 0.69 is obtained. In the second run, the fully-constrained network also having 20 degrees of freedom has a reference variance of 1.89. The a priori estimate for the reference variance in both adjustments is 1;

a. Is the minimally-constrained adjustment reference variance statistically equal to 1 at a 0.05 level of significance?  

b. Is the fully-constrained adjustment reference variance statistically equal to 1 at a 0.05 level of significance?

c. Are the two variances statistically equal at a 0.05 level of significance?

d. Is there statistical reason to be concerned about the presence of errors in either the control or the observations?

Answer 1

a)

Hypotheses.:

The null hypothesis and an alternative hypothesis. are given as,

Null hypothesis H₀: σ = 1

Alternative hypothesis H1: σ 1

Analysis plan.

The given significance level is 0.05.

Thus, the confidence interval is 95%.

Test:

The degree of freedom (DF) is calculated as,

DF = n - 1

20= n -1

Therefore, n=21

The test statistic (X²) is calculated as,

Now, P(Χ² > 13.8) = 0.84049 (Using Chi-Square Distribution Calculator)

Conclusion:

Since the P-value (0.84049) is greater than the significance level (0.05), we have to accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the minimally-constrained adjustment reference variance statistically equal to 1 at 0.05 level of significance.

b)

Hypotheses.:

The null hypothesis and an alternative hypothesis. are given as,

Null hypothesis H₀: σ = 1

Alternative hypothesis H1: σ 1

Analysis plan.

The given significance level is 0.05.

Thus, the confidence interval is 95%.

Test:

The degree of freedom (DF) is calculated as,

DF = n - 1

20= n -1

Therefore, n=21

The test statistic (X²) is calculated as,

Now, P(Χ² > 37.8) = 0.009367 (Using Chi-Square Distribution Calculator)

Conclusion:

Since the P-value (0.009367) is less than the significance level (0.05), we have to accept the null hypothesis.

From the above test we do not have sufficient evidence to the claim that the fully-constrained adjustment reference variance statistically equal to 1 at 0.05 level of significance.

c)

Hypotheses.:

The null hypothesis and an alternative hypothesis. are given as,

Null hypothesis H₀: σ_A² = σ_B²

Alternative hypothesis H_A: σ_A² σ_B²

Analysis plan.

The given significance level is 0.05.

Thus, the confidence interval is 95%.

Test:

The degree of freedom (DF) is calculated as,

DF₁ = n₁ - 1

20= n n₁ -1

Therefore, n₁=21

DF2 =n2 -1

20= n2 -1

Therefore, n2=21

The test statistic (F) is calculated as,

F = 0.36507

Since the first sample had the smaller standard deviation, this is a left-tailed test.

p value for the F distribution = 0.014568

Conclusion:

Since the P-value (0.014568) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test there is insufficient evidence to conclude that the two variances are statistically equal.