Question

2. [-/10 Points] DETAILS DEVORESTAT9 10.E.003. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER The lumen output was determined for each of I = 3 different brands of lightbulbs having the same wattage, with J = 7 bulbs of each brand tested. The sums of squares were computed as SSE = 4776.4 and SST = 598.4. State the hypotheses of interest (including word definitions of parameters). u; = sample average lumen output for brand i bulbs Ho: M1 = ll2 = 13 Ha: all three u;'s are unequal O M; = true average lumen output for brand i bulbs Ho: M1 M2 M3 Ha: at least two u;'s are equal OM; = sample average lumen output for brand i bulbs Ho: My & M2 # M3 Ha: all three u;'s are equal M; = true average lumen output for brand i bulbs Ho: M1 = M2 = M3 Ha: at least two j;'s are unequal Use the F test of ANOVA (a = 0.05) to decide whether there are any differences in true average lumen outputs among the three brands for this type of bulb by obtaining as much information as possible about the P-value. (Round your answer to two decimal places.) f= What can be said about the P-value for the test? OP-value > 0.100 0.050 < P-value < 0.100 O 0.010 < P-value < 0.050 O 0.001 < P-value < 0.010 OP-value < 0.001

Answer 1

The lumen output for each I=3 different brands of lightbulbs having the same wattage, with J=7 bulbs of each brand tested.

SSE = 4776.4

SSTr = 598.4

Hypotheses of interest are:

$\mu_i$ = true average lumen output for brand i bulbs

H₀: $\mu_1=\mu_2=\mu_3$

H_a: at least two $\mu_i$ 's are unequal.

Degrees of freedom for treatments = I - 1 = 3-1 = 2

Degrees of freedom for error = I( J-1 ) = 3*(7-1) = 3*6 = 18

We use F for ANOVA to test the above hypotheses.

Test statistic: $f=\frac{MSSTr}{MSSE}=\frac{SSTr/DFtr}{SSE/DFe}=\frac{598.4/2}{4776.4/18}=1.13$

Test statistic follows f distribution with ( 2,18 ) degrees of freedom

P-value = P[ f_2,18 > 71.84 ] = 0.34

P-value for this test is, p-value > 0.1

As the p-value > 0.05, we do not reject H₀ at 0.05 level of significance. Hence, there is no significant difference in the true average lumen outputs of three brands.

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