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an Dosen - Panduan Mahasiswa rencanaan Eksperimen (D) Statistik dan Perencanaan sement Minogue-10 Pertemuan 1 Bab 20 Acty has

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Answer #1

(a) The hypothesis being tested is:

H0: µ1 = µ2

H1: µ1 ≠ µ2

COMPANY A COMPANY A
2.6375 2.9 mean
0.5 0.6 std. dev.
8 12 n
-0.2625000 difference (COMPANY A - COMPANY A)
0.2474874 standard error of difference
0 hypothesized difference
-1.06 z
.2888 p-value (two-tailed)

The p-value is 0.2888.

Since the p-value (0.2888) is greater than the significance level (0.10), we cannot reject the null hypothesis.

Therefore, we can support the claim.

(b) The hypothesis being tested is:

H0: µ1 = µ2

H1: µ1 > µ2

COMPANY A COMPANY A
2.638 2.900 mean
0.625 0.895 std. dev.
8 12 n
17 df
-0.2625 difference (COMPANY A - COMPANY A)
0.3402 standard error of difference
0 hypothesized difference
-0.772 t
.7745 p-value (one-tailed, upper)

The p-value is 0.7745.

Since the p-value (0.7745) is greater than the significance level (0.10), we cannot reject the null hypothesis.

Therefore, we cannot support the claim.

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