A brand of powdered milk is advertised as having a net weight of 150 grams. A curious consumer obtained the net weight of 10 randomly selected cans. The values obtained are 256, 248, 242, 245, 246, 248, 250, 255, 243, and 149 grams. Is there a reason to believe that the average net weight of the powdered milk cans is less than 250 grams at 10% level of significance? Assume the net weight is normally distributed with unknown population variance.

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To determine if there is reason to believe that the average net weight of the powdered milk cans is less than 250 grams at the 10% level of significance, we will conduct a one-sample t-test.

Step 1: State the null and alternative hypotheses
The null hypothesis (H0) is that the average net weight of powdered milk cans is equal to 250 grams. The alternative hypothesis (H1) is that the average net weight is less than 250 grams.

H0: μ = 250
H1: μ < 250

Step 2: Calculate the test statistic
First, we need to calculate the sample mean and sample standard deviation.

Sample mean (x̄) = (256+248+242+245+246+248+250+255+243+149) / 10 = 2433 / 10 = 243.3 grams
Now, we find the sample variance:

Variance: Σ(xi−x̄)²/(N−1)
= (3.3^2 + 2^2 + 1.3^2 + 4.7^2 + 0^2) / 9
= (10.89 + 4 + 1.69 + 22.09 + 0) / 9
= 38.67 / 9
= 4.3

Sample standard deviation: √variance = √4.3 = 2.07

Now we can calculate the test statistic:

t = (x̄ - μ) / (s / √n)
t = (243.3 - 250) / (2.07 / √10)
t = -6.7 / (2.07 / 3.16)
t = -6.7 / 0.66
t ≈ -10.15

Step 3: Determine the critical value and p-value
We are given a 10% level of significance. Since this is a one-tailed test (we're only interested if the average weight is less than 250 grams), the critical value for a t-distribution with 9 degrees of freedom (sample size - 1) is -1.383 (rounded to three decimal places) [Reference: t-distribution table or calculator].

Using a t-table or calculator, we find the p-value associated with the test statistic -10.15. Given that our test statistic is much lower than -1.383, we can conclude that the p-value will be much smaller than 0.1 (10% level of significance).

Step 4: Compare the test statistic to the critical value and evaluate the p-value
Since our test statistic (-10.15) is less than the critical value (-1.383), we reject the null hypothesis. The p-value is also less than the level of significance (0.1), which further supports the rejection of the null hypothesis.

Conclusion:
There is reason to believe that the average net weight of the powdered milk cans is less than 250 grams at a 10% level of significance.