A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 402 gram setting. It is believed that the machine is underfilling the bags. A 23 bag sample had a mean of 400 grams with a variance of 900. A level of significance of 0.025 will be used. Assume the population distribution is approximately normal. Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

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Step-by-step explanation:

H0 : μ ≥ 402

H1 : μ < 402

Test statistic :

n = 23

Sample Mean, x = 400

Variance = 900

σ = sqrt(900) = 30

Test statistic = (x - μ) ÷ σ/sqrt(n)

Test statistic = (400 - 402) ÷ 30/sqrt(23)

Test statistic = - 2 / 6.2554324

Test statistic = - 0.3197221

Test statistic = - 0.3197

Critical value :

Using the Tcritical value calculator

Tcrit; α = 0.025, df = n - 1 = 23 - 1 = 22

Tcritical = 2.074

Reject Null : if Test statistic ≤ Tcritical (left tail test)

Since ;

Test statistic ≤ Tcritical ; We reject the Null

10.4 inches

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2x2x2.6

he kicked it up? is this the answer ?

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